5 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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33 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks

An icebox almost completely filled with ice at 0^°C is dipped into a large volume of water at 20^°C. The box has walls of surface area 2400cm², thickness 2.0mm and thermal conductivity $0.06\text{Wm}^{-1}{^{\circ}}\text{C}^{-1}.$ Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice $= 3.4\times10^5\text{Jkg}^{-1}.$

Answer

$\text{A}=2400\text{cm}^2=2400\times10^{-4}\text{m}^2$
$\ell=2\text{mm}=2\times10^{-3}\text{m}$
$\text{K}=0.06\text{w/m-}^\circ\text{C}$
$\theta_1=20^\circ\text{C}$
$\theta_2=0^\circ\text{C}$
$\frac{\text{Q}}{\text{t}}=\frac{\text{KA}(\theta_1-\theta_2)}{\ell}$
$=\frac{0.06\times2400\times10^{-4}\times20}{2\times10^{-3}}$
$=24\times6\times10^{-1}\times10$
$=24\times6=144\text{J/sec}.$
Rate in which ice melts $=\frac{\text{mL}_\text{f}}{\text{t}}$
$\Rightarrow\frac{\Delta\text{Q}}{\Delta\text{t}}=\Big(\frac{\text{m}}{\text{t}}\Big)\text{L}_\text{f}$
$\Rightarrow144=\Big(\frac{\text{m}}{\text{t}}\Big)\times3.4\times10^5$
$\Rightarrow\frac{\text{m}}{\text{t}}=\frac{144}{3.4\times10^{5}}\text{kg/s}$
$\Rightarrow\frac{\text{m}}{\text{t}}=\frac{144\times60\times60}{3.4\times10^5}\text{kg/h}$
$\Rightarrow\frac{\text{m}}{\text{t}}=1.52\text{kg/h}$

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Question 25 Marks

A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperatures. Assume that the emissivity of both the spheres is the same. Find the ratio of:

The rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere.
The rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the copper sphere. The specific heat capacity of aluminium = 900Jkg^-1°C^-1and that of copper= 390Jkg^-1°C^-1. The density of copper = 3.4 times the density of aluminium.

Answer

E → Energy radiated per unit area per unit time

Rate of heat flow → Energy radiated

Per time $=\text{E}\times\text{A}$

So, $\text{E}_\text{Al}=\frac{\text{e}\sigma\text{T}^4\times\text{A}}{\text{e}\sigma\text{T}^4\times\text{A}}$

$=\frac{4\pi\text{r}^2}{4\pi(2\text{r})^2}$

$=\frac{1}{4}$ $\big[\therefore1:4\big]$

Emissivity of both are same

$=\frac{\text{m}_1\text{S}_1\text{dT}_1}{\text{m}_2\text{S}_2\text{dT}_2}=1$

$\Rightarrow\frac{\text{dT}_1}{\text{dT}_2}=\frac{\text{m}_2\text{S}_2}{\text{m}_1\text{S}_1}$

$=\frac{\text{s}_14\pi\text{r}_1^3\times\text{S}_2}{\text{s}_24\pi\text{r}_2^3\times\text{S}_1}$

$=\frac{1\times\pi\times900}{3.4\times8\pi\times390}$

$=1:2:9$

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Question 35 Marks

Why does blowing over a spoonful of hot tea cools it? Does evaporation play a role? Does radiation play a role?

Answer

Here, major role is played by convection. When we blow air over a spoonful of hot tea, the air coming from our mouth has less temperature than the air above the tea. Since hot air has less density, it rises up and cool air goes down. In this way, the tea cools down.
We know that any hot body radiates.
So, the spoonful of tea will also radiate and as the temperature of the surrounding is less then the tea, the tea will cool down with time. Evaporation is also involved in this. On blowing over the hot tea, rate of evaporation increases and the cools down.

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Question 45 Marks

A pitcher with 1mm thick porous walls contains 10kg of water. Water comes to its outer surface and evaporates at the rate of 0.1gs^-1. The surface area of the pitcher (one side)= 200cm². The room temperature = 42^°C, latent heat of vaporization $ =2.27\times10^6\text{Jkg}^{-1},$ and the thermal conductivity of the porous walls $=0.80\text{Js}^{-1}\text{m}^{-1}{^{\circ}}\text{C}^{-1}.$ Calculate the temperature of water in the pitcher when it attains a constant value.

Answer

$\ell=1\text{mm}=10^{-3}\text{m},\ \text{m}=10\text{kg}$
$\text{A}=200\text{cm}^2=2\times10^{-2}\text{m}^2$
$\text{L}_\text{vap}=2.27\times10^6\text{J/kg}$
$\text{K}=0.80\text{J/m-s-}^\circ\text{C}$
$\text{dQ}=2.27\times10^6\times10,$
$\frac{\text{dQ}}{\text{dt}}=\frac{2.27\times10^7}{10^5}$
$=2.27\times10^2\text{J/s}$
Again we know
$\frac{\text{dQ}}{\text{dt}}=\frac{0.80\times2\times10^{-2}\times(42-\text{T})}{1\times10^{-3}}$
So, $\frac{8\times2\times10^{-3}(42-\text{T})}{10^{-3}}=2.27\times10^2$
$\Rightarrow16\times42-16\text{T}$
$\Rightarrow\text{T}=27.8\approx28^\circ\text{C}$

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Question 55 Marks

Find the rate of heat flow through a cross section of the rod shown in figure, $\big(\theta_2>\theta_1\big).$ Thermal conductivity of the material of the rod is K.

Answer

$\phi=\frac{\text{r}_2-\text{r}_1}{\text{L}}=\frac{(\text{y}-\text{r}_1)}{\text{x}}$
$\Rightarrow\text{x}\text{r}_2-\text{x}\text{r}_1=\text{yL}-\text{r}_1\text{L}$
Differentiating wr to ‘x’
$\Rightarrow\text{r}+2\text{r}_1=\frac{\text{Ldy}}{\text{dx}}-0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{r}_2-\text{r}_1}{\text{L}}$
$\Rightarrow\text{dx}=\frac{\text{dyL}}{(\text{r}_2-\text{r}_1)}\ \dots(1)$

Now $\frac{\text{Q}}{\text{T}}=\frac{\text{K}\pi\text{y}^2\text{d}\theta}{\text{dx}}$
$\Rightarrow\frac{\theta\text{dx}}{\text{T}}=\text{K}\pi\text{y}^2\text{d}\theta$
$\Rightarrow\frac{\theta\text{Ldy}}{\text{r}_2\text{r}_1}=\text{K}\pi\text{y}^2\text{d}\theta$ from(1)
$\Rightarrow\text{d}\theta\frac{\text{QLdy}}{(\text{r}_2-\text{r}_1)\text{K}\pi\text{y}^2}$
Integrating both side
$\Rightarrow\int\limits_{\theta_1}^{\theta_2}\text{d}\theta=\frac{\text{QL}}{(\text{r}_2-\text{r}_1)\text{K}\pi}\int\limits^{\text{r}_2}_{\text{r}_1}\frac{\text{dy}}{\text{y}}$
$\Rightarrow(\theta_2-\theta_1)=\frac{\text{QL}}{(\text{r}_2-\text{r}_1)}\times\Big[\frac{-1}{\text{y}}\Big]^{\text{r}_2}_{\text{r}_1}$
$\Rightarrow(\theta_2-\theta_1)=\frac{\text{QL}}{(\text{r}_2-\text{r}_1)\text{K}\pi}\times\Big[\frac{1}{\text{r}_1}-\frac{1}{\text{r}_2}\Big]$
$\Rightarrow(\theta_2-\theta_1)=\frac{\text{QL}}{(\text{r}_2-\text{r}_1)\text{K}\pi}\times\Big[\frac{\text{r}_2-\text{r}_1}{\text{r}_1+\text{r}_2}\Big]$
$\Rightarrow\text{Q}=\frac{\text{K}\pi\text{r}_1\text{r}_2(\theta_2-\theta_1)}{\text{L}}$

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Question 65 Marks

A calorimeter of negligible heat capacity contains 100cc of water at 40^°C. The water cools to 35^°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40^°C. Find the time taken for the temperature to become 35^°C under similar conditions. Specific heat capacities of water and K-oil are 4200Jkg^-1K^-1 and 2100Jkg^-1K^-1 respectively. Density of K-oil = 800kgm³.

Answer

$\text{v}=100\text{cc}$
$\triangle\theta=5^\circ\text{C}$
$\text{t}=5\text{ min}$
For water
$\frac{\text{mS}\triangle\theta}{\text{dt}}=\frac{\text{KA}}{\text{l}}\triangle\theta$
$\Rightarrow\frac{100\times10^{-3}\times1000\times4200}{5}=\frac{\text{KA}}{\text{l}}$
For Kerosene
$\frac{\text{ms}}{\text{at}}=\frac{\text{KA}}{\text{l}}$
$\Rightarrow\frac{100\times10^{-3}\times800\times2100}{\text{t}}=\frac{\text{KA}}{\text{l}}$
$\Rightarrow\frac{100\times10^{-3}\times800\times2100}{\text{t}}=\frac{100\times10^{-3}\times1000\times4200}{5}$
$\Rightarrow\text{T}=\frac{5\times800\times2100}{1000\times4200}=2\text{ min}$

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Question 75 Marks

Two identical metal balls one at $\text{T}_1=300\text{K}$ and the other at $\text{T}_2=600\text{K}$ are kept at a distance of 1m in vacuum. Will the temperatures equalise by radiation? Will the rate of heat gained by the colder sphere be proportional to $\text{T}_2^4-\text{T}_1^4$ as may be expected from the Stefan's law?

Answer

Yes, the temperature of the balls can be equalised by radiation. This is because both the spheres will emit radiations in all the directions at different rates.
The ball kept at the temperature of 300K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of 600K. Also, it losses energy by radiation.
Similarly, the ball kept at the temperature of 600K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of 600K. Also, it losses energy by radiation.
A time comes when the temperature of both the bodies becomes equal.
Yes, the rate of heat gained by the colder sphere is proportional to $\text{T}_2^4-\text{T}_1^4$

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Question 85 Marks

Two bodies of masses m₁ and m₂ and specific heat capacities s₁ and s₂ are connected by a rod of length l, cross-sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time t = 0, the temperature of the first body is T₁ and the temperature of the second body is T₂ (T₂ > T₁). Find the temperature difference between the two bodies at time t.

Answer

$\frac{\text{Q}}{\text{t}}=\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{L}}$
Rise in Temp. in $\text{T}_2\Rightarrow\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lm}_1\text{s}_1}$ Fall in Temp in $\text{T}_1\Rightarrow\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lm}_2\text{s}_2}$
Final Temp. $\text{T}_1=\text{T}_1-\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lm}_1\text{s}_1}$ Final Temp. $\text{T}_2=\text{T}_2+\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lm}_1\text{s}_1}$
$\Rightarrow\frac{\triangle\text{T}}{\text{dt}}=\text{T}_1-\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lm}_1\text{s}_1}-\text{T}_2-\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lm}_2\text{s}_2}\\ =(\text{T}_1-\text{T}_2)-\Big[\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lm}_1\text{s}_1}+\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lm}_2\text{s}_2}\Big]$
$\Rightarrow\frac{\text{dT}}{\text{dt}}=\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{L}}\Big(\frac{1}{\text{m}_1\text{s}_1}+\frac{1}{\text{m}_2\text{s}_2}\Big)$
$\Rightarrow\frac{\text{dT}}{(\text{T}_1-\text{T}_2)}=\frac{\text{KA}}{\text{L}}\Big(\frac{\text{m}_2\text{s}_2+\text{m}_1\text{s}_2}{\text{m}_1\text{s}_1\text{m}_2\text{s}_2}\Big)\text{dt}$
$\Rightarrow\text{ln}\triangle\text{t}=-\frac{\text{KA}}{\text{L}}\Big(\frac{\text{m}_2\text{s}_2+\text{m}_1\text{s}_2}{\text{m}_1\text{s}_1\text{m}_2\text{s}_2}\Big)\text{t}+\text{C}$
At time $\text{t}=0,\ \text{T}=\text{T}_0,\ \triangle\text{T}=\triangle\text{T}_0$ $\Rightarrow\text{C}=\text{ln}\triangle\text{T}_0$
$\Rightarrow\text{ln}\frac{\triangle\text{T}}{\triangle\text{T}_0}=-\frac{\text{KA}}{\text{L}}\Big(\frac{\text{m}_2\text{s}_2+\text{m}_1\text{s}_2}{\text{m}_1\text{s}_1\text{m}_2\text{s}_2}\Big)\text{t}$
$\Rightarrow\frac{\triangle\text{T}}{\triangle\text{T}_0}=\text{e}^{\frac{\text{KA}}{\text{L}}\Big(\frac{\text{m}_2\text{s}_2+\text{m}_1\text{s}_2}{\text{m}_1\text{s}_1\text{m}_2\text{s}_2}\Big)\text{t}}$
$\Rightarrow\triangle\text{T}=\triangle\text{T}_0\text{e}^{\frac{\text{KA}}{\text{L}}\Big(\frac{\text{m}_2\text{s}_2+\text{m}_1\text{s}_2}{\text{m}_1\text{s}_1\text{m}_2\text{s}_2}\Big)\text{t}}\\=(\text{T}_2-\text{T}_1)\text{e}^{\frac{\text{KA}}{\text{L}}\Big(\frac{\text{m}_2\text{s}_2+\text{m}_1\text{s}_2}{\text{m}_1\text{s}_1\text{m}_2\text{s}_2}\Big)\text{t}}$

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Question 95 Marks

The left end of a copper rod (length = 20cm area of cross section = 0.20cm) is maintained at 20^°C and the right end is maintained at 80^°C. Neglecting any loss of heat through radiation, find,

The temperature at a point 11cm from the left end
The heat current through the rod. Thermal conductivity of copper $=385\text{Wm}^{-1}{^{\circ}}\text{C}^{-1}.$

Answer

$\ell=20\text{cm}=20\times10^{-2}\text{m}$

$\text{A}=0.2\text{cm}^2=0.2\times10^{-4}\text{m}^2$

$\theta_1=\text{80}^\circ\text{C},\ \theta_2-20^\circ\text{C},\ \text{K}=385$

$\frac{\text{Q}}{\text{t}}=\frac{\text{KA}(\theta_1-\theta_2)}{\ell}$

$=\frac{385\times0.2\times10^{-4}(80-20)}{20\times10^{-2}}$

$=385\times6\times10^{-4}\times10$

$=2310\times10^{-3}=2.31$

Let the temp of the 11cm point be $\theta$

$\frac{\triangle\theta}{\triangle\text{l}}=\frac{\text{Q}}{\text{tKA}}$

$\Rightarrow\frac{\triangle\theta}{\triangle\text{l}}=\frac{2.31}{385\times0.2\times10^{-4}}$

$\Rightarrow\frac{\theta-20}{11\times10^{-2}}=\frac{2.31}{385\times0.2\times10^{-4}}$

$\Rightarrow\theta-20=\frac{2.31\times10^4}{385\times0.2}\times11\times10^{-2}=33$

$\Rightarrow\theta=33+20=53^\circ\text{C}$

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Question 105 Marks

Water at 50^°C is filled in a closed cylindrical vessel of height 10cm and cross sectional area 10cm². The walls of the vessel are adiabatic but the flat parts are made of 1mm thick aluminium $(\text{K}=200\text{Js}^{-1}\text{m}^{-1}{^{\circ}}\text{C}^{-1}).$ Assume that the outside temperature is 20^°C. The density of water is 1000kgm^-3, and the specific heat capacity of water $= 4200\text{Jkg}^{-1}{^{\circ}}\text{C}^{-1}.$ Estimate the time taken for the temperature to fall by 1.0^°C. Make any simplifying assumptions you need but specify them.

Answer

$\text{A}=10\text{cm}^2,\ \text{h}=10\text{cm}$
$\frac{\triangle\text{Q}}{\triangle\text{t}}=\frac{\text{KA}(\theta_1-\theta_2)}{\ell}$
$=\frac{200\times10^{-3}\times30}{1\times10^{-3}}$
$=6000$
Since heat goes out from both surfaces. Hence net heat coming out.
$=\frac{\triangle\text{Q}}{\triangle\text{t}}=6000\times2=12000,$ $\Big[\frac{\triangle\text{Q}}{\triangle\text{t}}=\text{Ms}\frac{\triangle\theta}{\theta\text{t}}\Big]$
$\Rightarrow6000\times2=10^{-3}\times10^{-1}1000\times4200\times\frac{\triangle\theta}{\triangle\text{t}}$
$\Rightarrow\frac{\triangle\theta}{\triangle\text{t}}=\frac{72000}{420}$
$=28.57$
So, in 1Sec. 28.57^°C is dropped
Hence for drop of $1^\circ\text{C}\frac{1}{28.57}\text{sec}.$
$=0.0.35\text{sec}.$ is required.

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Question 115 Marks

A metal block of heat capacity 80J^°C^-1 placed in a room at 20^°C is heated electrically. The heater is switched off when the temperature reaches 30°C. The temperature of the block rises at the rate of 2^°Cs^-1 just after the heater is switched on and falls at the rate of 0.2^°Cs^-1 just after the heater is switched off. Assume Newton's law of cooling to hold.

Find the power of the heater.
Find the power radiated by the block just after the heater is switched off.
Find the power radiated by the block when the temperature of the block is 25^°C.
Assuming that the power radiated at 25^°C represents the average value in the heating process, find the time for which the heater was kept on.

Answer

Given: Heat capacity = m × s = 80J/^°C

$\Big(\frac{\text{d}\theta}{\text{dt}}\Big)_\text{increase}=2^\circ\text{C/s}$

$\Big(\frac{\text{d}\theta}{\text{dt}}\Big)_\text{decrease}=0.2^\circ\text{C/s}$

Power of heater $=\text{mS}\Big(\frac{\text{d}\theta}{\text{dt}}\Big)_\text{increasing}=80\times2=160\text{W}$
Power radiated $=\text{mS}\Big(\frac{\text{d}\theta}{\text{dt}}\Big)_\text{decreasing}=80\times0.2=16\text{W}$
Now $\text{mS}\Big(\frac{\text{d}\theta}{\text{dt}}\Big)_\text{decreasing}=\text{K}(\text{T}-\text{T}_0)$

$\Rightarrow16=\text{K}(30-20)$

$\Rightarrow\text{K}=\frac{16}{10}=1.6$

Now, $\frac{\text{d}\theta}{\text{dt}}=\text{K}(\text{T}-\text{T}_0)$

$=1.6\times(30-25)=1.6\times5$

$=8\text{W}$

$\text{P.t}=\text{H}$

$\Rightarrow8\times\text{t}$

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Question 125 Marks

A hollow tube has a length l₁ inner radius R₁ and outer radius R₂. The material has a thermal conductivity K. Find the heat flowing through the walls of the tube if.

The flat ends are maintained at temperatures T₁ and T₂(T₂ > T₁)
The inside of the tube is maintained at temperature T₁ and the outside is maintained at T₂.

Answer

$\text{T}_1>\text{T}_2$
$\text{A}=\pi\big(\text{R}_2^2-\text{R}_1^2\big)$
So. $\text{Q}=\frac{\text{KA}(\text{T}_2-\text{T}_1)}{\text{l}}$
$=\frac{\text{KA}\big(\text{R}_2^2-\text{R}_1^2\big)(\text{T}_2-\text{T}_1)}{\text{l}}$
Considering a concentric cylindrical shell of radius ‘r’ and thickness ‘dr’. The radial heat flow through the shell

$\text{H}=\frac{\text{dQ}}{\text{dt}}=-\text{KA}\frac{\text{d}\theta}{\text{dt}}$ [(-)ve because as r - increases $\theta$ decreases]
$\text{A}=2\pi\text{rl},\ \text{H}=-2\pi\text{rl}\text{K}\frac{\text{d}\theta}{\text{dt}}$
$\Rightarrow\int\limits_{\text{R}_1}^{\text{R}_2}\frac{\text{dr}}{\text{r}}$
$=-\frac{2\pi\text{LK}}{\text{H}}\int\limits^{\text{T}_2}_{\text{T}_1}\text{d}\theta$
Integrating and simplifying we get
$\text{H}=\frac{\text{dQ}}{\text{dt}}$
$=\frac{2\pi\text{KL}(\text{T}_2-\text{T}_1)}{\text{Loge}\Big(\frac{\text{R}_2}{\text{R}_1}\Big)}$
$=\frac{2\pi\text{KL}(\text{T}_2-\text{T}_1)}{\text{ln}\Big(\frac{\text{R}_2}{\text{R}_1}\Big)}$

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Question 135 Marks

Figure, shows two adiabatic vessels, each containing a mass m of water at different temperatures. The ends of a metal rod of length L, area of cross section A and thermal conductivity K, are inserted in the water as shown in the figure. Find the time taken for the difference between the temperatures in the vessels to become half of the original value. The specific heat capacity of water is s. Neglect the heat capacity of the rod and the container and any loss of heat to the atmosphere.

Answer

$\frac{\text{Q}}{\text{t}}=\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{L}}$
Rise in Temp. in $\text{T}_2\Rightarrow\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lms}}$ Fall in Temp in $\text{T}_1=\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lms}}$
Final Temp. $\text{T}_1\Rightarrow\text{T}_1-\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lms}}$ Final Temp $\text{T}_2=\text{T}_2+\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lms}}$
Final $\frac{\triangle\text{T}}{\text{dt}}=\text{T}_1-\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lms}}-\text{T}_2-\frac{\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lms}}$
$=(\text{T}_1-\text{T}_2)-\frac{2\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lms}}=\frac{\text{dT}}{\text{dt}}$
$=-\frac{2\text{KA}(\text{T}_1-\text{T}_2)}{\text{Lms}}$
$\Rightarrow\int\limits^{(\text{T}_1-\text{T}_2)}_{(\text{T}_1-\text{T}_2)}\frac{\text{dt}}{(\text{T}_1-\text{T}_2)}=\frac{-2\text{KA}}{\text{Lms}}\text{dt}$
$\Rightarrow\text{ln}\frac{\frac{(\text{T}_1-\text{T}_2)}{2}}{(\text{T}_1-\text{T}_2)}=\frac{-2\text{KAt}}{\text{Lms}}$
$\Rightarrow\text{ln}\Big(\frac{1}{2}\Big)=\frac{-2\text{KAt}}{\text{Lms}}$
$\Rightarrow\text{ln}_2=\frac{2\text{KAt}}{\text{Lms}}$
$\Rightarrow\text{t}=\text{ln}_2\frac{\text{lms}}{2\text{KA}}$

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Question 145 Marks

A hot body placed in a surrounding of temperature $\theta_0$ obeys Newton's law of cooling $\frac{\text{d}\theta}{\text{dt}}=-\text{k}(\theta-\theta_0).$ Its temperature at t = 0 is $\theta_1.$ The specific heat capacity of the body is s and its mass is m. Find,

The maximum heat that the body can lose.
The time starting from t = 0 in which it will lose 90% of this maximum heat.

Answer

$\frac{\text{d}\theta}{\text{dt}}=-\text{K}(\text{T}-\text{T}_0)$

Temp. at t = 0 is $\theta_1$

Max. Heat that the body can loose $\triangle\text{Q}_\text{m}=\text{ms}(\theta_1-\theta_2)$

$\big(\therefore\text{as},\ \triangle\text{t}=\theta_1-\theta_0\big)$

If the body loses 90% of the max heat the decrease in its temp. will be

$\frac{\triangle\text{Q}_\text{m}\times9}{10\text{ms}}=\frac{(\theta_1-\theta_0)\times9}{10}$

If it takes time t₁, for this process, the temp. at t₁

$=\theta_1-(\theta_1-\theta_0)\frac{9}{10}$

$=\frac{10\theta_1-9\theta_1-9\theta_0}{10}$

$=\frac{\theta_1-9\theta_0}{10}\times1\ \dots(1)$

Now, $\frac{\text{d}\theta}{\text{dt}}=-\text{K}(\theta-\theta_1)$

Let $\theta=\theta_1$ at $\text{t}=0;$ and $\theta$ be temp. at time t

$\int\limits^\theta_{\theta_1}\frac{\text{d}\theta}{\theta_1-\theta}=-\text{K}\int\limits_0^\text{t}\text{dt}$

$\text{ln}\frac{\theta-\theta_0}{\theta_1-\theta_0}=-\text{Kt}$

$\theta-\theta_0=(\theta_1-\theta_0)\text{e}^{-\text{Kt}}\ \dots(2)$

Putting value in the Eq. (1) and Eq. (2)

$\frac{\theta_1-9\theta_0}{10}-\theta_0=(\theta_1-\theta_0)\text{e}^{-\text{Kt}}$

$\Rightarrow\text{t}_1=\frac{\text{ln}10}{\text{k}}$

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Question 155 Marks

A rod of negligible heat cafacity has length 20cm, area of cross section 1.0cm and thermal conductivity 200Wm^-1°C^-1. The temperature of one end is maintained at 0^°C and that of the other end is slowly and linearly varied from 0^°C to 60^°C in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes.

Answer

$\frac{\text{d}\theta}{\text{dt}}=\frac{60}{10\times60}=0.1^\circ\text{C/sec}$
$\frac{\text{dQ}}{\text{dt}}=\frac{\text{Ka}}{\text{d}}(\theta_1-\theta_2)$
$=\frac{\text{KA}\times0.1}{\text{d}}+\frac{\text{KA}\times0.2}{\text{d}}+_{\dots}+\frac{\text{KA}\times60}{\text{d}}$
$=\frac{\text{KA}}{\text{d}}(0.1+0.2+_{\dots}+60)$
$=\frac{\text{KB}}{\text{d}}\times\frac{600}{2}\times(2\times0.1+599\times0.1)$
$\Big[\therefore\text{a}+2\text{a}+_{\dots}+\ \text{na}=\frac{\text{n}}{2}\big\{2\text{a}+(\text{n}-1)\text{a}\big\}\Big]$
$=\frac{200\times1\times10^{-4}}{20\times10^{-2}}\times300\times(0.2+59.9)$
$=\frac{200\times10^{-2}\times300\times60.1}{20}$
$=3\times10\times60.1$
$=1803\text{w}\approx1800\text{w}$

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Question 165 Marks

The two rods shown in figure, have identical geometrical dimensions. They are in contact with two heat baths at temperatures 100^°C and 0^°C. The temperature of the junction is 70^°C. Find the temperature of the junction if the rods are interchanged.

Answer

Now, $\frac{\text{Q}}{\text{t}} $ remains same in both cases

In case I: $\frac{\text{K}_\text{A}\text{XA}\times(100-70)}{\ell}=\frac{\text{K}_\text{B}\times\text{A}\times(70-0)}{\ell}$
$\Rightarrow30\text{K}_\text{A}=70\text{K}_\text{B}$

In case II: $\frac{\text{K}_\text{B}\text{XA}\times(100-70)}{\ell}=\frac{\text{K}_\text{A}\times\text{A}\times(\theta-0)}{\ell}$
$\Rightarrow100\text{K}_\text{B}-\text{K}_\text{B}\theta=\text{K}_\text{A}\theta$
$\Rightarrow100\text{K}_\text{B}-\text{K}_\text{B}\theta=\frac{70}{30}\text{K}_\text{B}\theta$
$\Rightarrow100=\frac{7}{3}\theta+\theta$
$\Rightarrow\theta=\frac{300}{10}=30^\circ\text{C}$

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Question 175 Marks

A metal ball of mass 1kg is heated by means of a 20W heater in a room at 20^°C. The temperature of the ball becomes steady at 50^°C.

Find the rate of loss of heat to the surrounding when the ball is at 50^°C.
Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball is at 30^°C.
Assume that the temperature of the ball rises uniformly from 20^°c to 30^°C in 5 minutes. Find the total loss of heat to the surrounding during this period.
Calculate the specific heat capacity of the metal.

Answer

In steady state condition as no heat is absorbed, the rate of loss of heat by conduction is equal to that of the supplied.

i.e., H = P

m = 1Kg, Power of Heater = 20W, Room Temp. = 20^°C

$\text{H}=\frac{\text{d}\theta}{\text{dt}}$

$\text{P}=20\text{ Watt}$

By Newton’s law of cooling

$\frac{-\text{d}\theta}{\text{dt}}=\text{K}(\theta-\theta_0)$

$-20=\text{K}(50-20)$

$\Rightarrow\text{K}=\frac{2}{3}$

Again, $\frac{-\text{d}\theta}{\text{dt}}=\text{K}(\theta-\theta_0)$

$=\frac{2}{3}\times(30-20)$

$=\frac{20}{3}\text{w}$

$\Big(\frac{\text{dQ}}{\text{dt}}\Big)_{20}=0,\ \Big(\frac{\text{dQ}}{\text{dt}}\Big)_{30}=\frac{20}{3},\ \Big(\frac{\text{dQ}}{\text{dt}}\Big)_\text{avg}=\frac{10}{3}$

$\text{T}=5\text{ min}=300$

Heat liberated $=\frac{10}{3}\times300=1000\text{J}$

Net Heat absorbed = Heat supplied - Heat Radiated

= 6000 - 1000 = 5000J

Now, $\text{m}\triangle\theta'=5000$

$\Rightarrow\text{S}=\frac{5000}{\text{m}\triangle\theta}$

$=\frac{5000}{1\times10}$

$=5000\text{Jkg}^{-1}{^{\circ}}\text{C}^{-1}$

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Question 185 Marks

A hollow metallic sphere of radius 20cm surrounds a concentric metallic sphere of radius 5cm. The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at 50^°C and 10^°C respectively and it is found that 100J of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres.

Answer

$\text{a}=\text{r}_1=5\text{cm}=0.05\text{m}$
$\text{b}=\text{r}_2=20\text{cm}=0.2\text{m}$
$\theta_1=\text{T}_1=50^\circ\text{C},\ \theta_2=\text{T}_2=10^\circ\text{C}$
Now, considering a small strip of thickness ‘dr’ at a distance ‘r’

$\text{A}=4\pi\text{r}^2$
$\text{H}=-4\pi\text{r}^2\text{K}\frac{\text{d}\theta}{\text{dr}}$ [(-)ve because with increase of r, $\theta$ decreases]
$=\int\limits_\text{a}^\text{b}\frac{\text{dr}}{\text{r}^2}=\frac{-4\pi\text{K}}{\text{H}}\int\limits^{\theta_2}_{\theta_1}\text{d}\theta$ On integration
$\text{H}=\frac{\text{dQ}}{\text{dt}}=\text{K}\frac{4\pi\text{ab}(\theta_1-\theta_2)}{(\text{b}-\text{a})}$
Putting the values we get
$\frac{\text{K}\times4\times3.14\times5\times20\times40\times10^{-3}}{15\times10^{-2}}=100$
$\Rightarrow\text{K}=\frac{15}{4\times3.14\times4\times10^{-1}}$
$=2.985\approx3\text{w/m-}^\circ\text{C}$

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Question 195 Marks

A metal rod of cross sectional area 1.0cm² is being heated at one end. At one time, the temperature gradient is 5.0^°C/cm^-1 at cross section A and is 2.5^°C/cm^-1 at cross section B. Calculate the rate at which the temperature is increasing in the part AB of the rod. The heat capacity of the part AB = 0.40J^°C^-1, thermal conductivity of the material of the rod = 200Wm^-1°C^-1. Neglect any loss of heat to the atmosphere.

Answer

The rate of heat flow per sec.
$=\frac{\text{dQ}_\text{A}}{\text{dt}}=\text{KA}\frac{\text{d}\theta}{\text{dt}}$
The rate of heat flow per sec.
$=\frac{\text{dQ}_\text{A}}{\text{dt}}=\text{KA}\frac{\text{d}\theta}{\text{dt}}$
The rate of heat flow per sec.
$=\frac{\text{dQ}_\text{B}}{\text{dt}}=\text{KA}\frac{\text{d}\theta_\text{B}}{\text{dt}}$
This part of heat is absorbed by the red.
$=\frac{\text{Q}}{\text{t}}=\frac{\text{ms}\triangle\theta}{\text{dt}}$
Where $\frac{\text{d}\theta}{\text{dt}}$ = Rate of net temp. variation
$\Rightarrow\frac{\text{msd}\theta}{\text{dt}}=\text{KA}\frac{\text{d}\theta_\text{A}}{\text{dt}}-\text{KA}\frac{\text{d}\theta_\text{B}}{\text{dt}}$
$\Rightarrow\text{ms}\frac{\text{d}\theta}{\text{dt}}=\text{KA}\Big(\frac{\text{d}\theta_\text{A}}{\text{dt}}-\frac{\text{d}\theta_\text{B}}{\text{dt}}\Big)$
$\Rightarrow0.4\times\frac{\text{d}\theta}{\text{dt}}=200\times1\times10^{-4}(5-2.5)^\circ\text{C/cm}$
$\Rightarrow0.4\times\frac{\text{d}\theta}{\text{dt}}=200\times10^{-4}\times2.5$
$\Rightarrow\frac{\text{d}\theta}{\text{dt}}=\frac{200\times2.5\times10^{-4}}{0.4\times10^{-2}}\ {^\circ\text{C/m}}=1250\times10^{-2}=12.5^\circ\text{C/m}$

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Question 205 Marks

A spherical tungsten piece of radius 1.0cm is suspended in an evacuated chamber maintained at 300K. The piece is maintained at 1000K by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is 0.30 and the Stefan constant $\sigma$ is 6.0 × 10^-8Wm^-2K^-4.

Answer

$\text{r}=1\text{cm}=1\times10^{-3}\text{m}$
$\text{A}=4\pi(10^{-2})^2$
$=4\pi\times10^{-4}\text{m}^2$
$\text{E}=0.3,\ \sigma=6\times10^{-8}$
$\frac{\text{E}}{\text{t}}=\text{A}\sigma\text{e}\big(\text{T}^4_1-\text{T}^4_2)$
$=0.3\times6\times10^{-8}\times4\pi\times10^{-4}\times\big[(100)^4-(300)^4\big]$
$=0.3\times6\times4\pi\times10^{-12}\times\big[1-0.0081\big]\times10^{-12}$
$=0.3\times6\times4\times3.14\times9919\times10^{-4}$
$=4\times10\times3.14\times9919\times10^{-5}$
$=22.4\approx22\text{W}$

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Question 215 Marks

A body cools down from 50^°C to 45^°C in 5 minutes and to 40^°C in another 8 minutes. Find the temperature of the surrounding.

Answer

$50^\circ\text{C},\ 45^\circ\text{C},\ 40^\circ\text{C}$
Let the surrounding temperature be ‘T’^°C
Avg. $\text{t}=\frac{50+45}{2}=47.5$
Avg. temp. diff. from surrounding $\text{T}=47.5-\text{T}$
Rate of fall of temp $=\frac{50-45}{5}=1^\circ\text{C/mm}$
From Newton’s Law $1^\circ\text{C/mm}=\text{bA}\times\text{t}$
$\Rightarrow\text{bA}=\frac{1}{\text{t}}$
$=\frac{1}{47.5-\text{T}}\ \dots(1)$
In second case,
Avg, temp $=\frac{40+45}{2}=42.5$
Avg. temp. diff. from surrounding $\text{t}'=42.5-\text{t}$
Rate of fall of temp $=\frac{45-40}{8}=\frac{5}{8}\ ^\circ\text{C/mm}$
From Newton’s Law $\frac{5}{\text{B}}=\text{bAt}'$
$\Rightarrow\frac{5}{8}=\frac{1}{(47.5-\text{T})}\times(42.5-\text{T})$
By C & D [Componendo & Dividendo method]
We find, $\text{T}=34.1^\circ\text{C}$

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Question 225 Marks

Steam at 120^°C is continuously passed through a 50cm long rubber tube of inner and outer radii 1.0cm and 1.2cm. The room temperature is 30^°C. Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber = 0.15Js^-1m^-1^°C^-1.

Answer

Given, $\text{K}_\text{rubber}=0.15\text{J/m-s-}^\circ\text{C},\ \text{T}_2-\text{T}_1=90^\circ\text{C}$

We know for radial conduction in a Cylinder

$\frac{\text{Q}}{\text{t}}=\frac{2\pi\text{Kl}(\text{T}_2-\text{T}_1)}{\text{ln}\Big(\frac{\text{R}_2}{\text{R}_1}\Big)}$

$=\frac{2\times3.14\times15\times10^{-2}\times50\times10^{-1}\times90}{\text{ln}\Big(\frac{1.2}{1}\Big)}$

$=232.5\approx233\text{J/s}.$

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Question 235 Marks

A calorimeter contains 50g of water at 50^°C. The temperature falls to 45^°C in 10 minutes. When the calorimeter contains 100g of water·at 50^°0, it takes 18 minutes for the temperature to become 45^°C. Find the water equivalent of the calorimeter.

Answer

Let the water eq. of calorimeter = m
$\frac{(\text{m}+50\times10^{-3})\times4200\times5}{10}=$ Rate of heat flow
$\frac{(\text{m}+50\times10^{-3})\times4200\times5}{18}=$ Rate of flow
$\Rightarrow\frac{(\text{m}+50\times10^{-3})\times4200\times5}{10}=\frac{(\text{m}+50\times10^{-3})\times4200\times5}{18}$
$\Rightarrow(\text{m}+50\times10^{-3})18=10\text{m}+1000\times10^{-3}$
$\Rightarrow18\text{m}+18\times50\times10^{-3}=10\text{m}+1000\times10^{-3}$
$\Rightarrow8\text{m}=100\times10^{-3}\text{kg}$
$\Rightarrow\text{m}=12.5\times10^{-3}\text{kg}=12.5\text{g}$

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Question 245 Marks

A steel frame $(\text{K}=45\text{Wm}^{-1}{^{\circ}}\text{C}^{-1})$of total length 60cm and cross sectional area 0.20cm², forms three sides of a square. The free ends are maintained at 20^°C and 40^°C. Find the rate of heat flow through a cross section of the frame.

Answer

$\text{K}=45\text{w/m-}^\circ\text{C}$

$\ell=60\text{cm}=60\times10^{-2}\text{m}$

$\text{A}=0.2\text{cm}^2=0.2\times10^{-4}\text{m}^2$

Rate of heat flow,

$=\frac{\text{KA}(\theta_1-\theta_2)}{\ell}$

$=\frac{45\times0.2\times10^{-4}\times20}{60\times10^{-2}}$

$=30\times10^{-3}=0.03\text{w}$

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Question 255 Marks

Figure, shows water in a container having 2.0mm thick walls made of a material of thermal conductivity $0.50\text{Wm}^{-1}{^{\circ}}\text{C}^{-1}.$ The container is kept in a melting-ice bath at 0^°C. The total surface area in contact with water is 0.05m². A wheel is clamped inside the water and is coupled to a block of mass M as shown in the figure. As the block goes down, the wheel rotates. It is found that after some time a steady state is reached in which the block goes down with a constant speed of 10cms^-1 and the temperature of the water remains constant at 1.0^°C. Find the mass M of the block. Assume that the heat flows out of the water only through the walls in contact. Take g = 10ms^-2.

Answer

Given $\theta_1=1^\circ\text{C},\ \theta_2=0^\circ\text{C}$

$\text{K}=0.50\text{w/m-}^\circ\text{C},\ \text{d}=2\text{mm}=2\times10^{-3}\text{m}$

$\text{A}=5\times10^{-2}\text{m}^2,\ \text{v}=10\text{cm/s}=0.1\text{m/s}$

Power = Force × Velocity = Mg × v

Again Power $=\frac{\text{dQ}}{\text{dt}}=\frac{\text{KA}(\theta_1-\theta_2)}{\text{d}}$

So, Mgv $=\frac{\text{KA}(\theta_1-\theta_2)}{\text{d}}$

$\Rightarrow\text{M}=\frac{\text{KA}(\theta_1-\theta_2)}{\text{dvg}}$

$=\frac{5\times10^{-1}\times5\times^{-2}\times1}{2\times10^{-3}\times10^{-1}\times10}$

$=12.5\text{kg}.$

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Question 265 Marks

A spherical ball of surface area 20cm² absorbs any radiation that falls on it. It is suspended in a closed box maintained at 57^°C.

Find the amount of radiation falling on the ball per second.
Find the net rate of heat flow to or from the ball at an instant when its temperature is 200^°C. Stefan constant = 6.0 × 10^-8Wm^-2K^-4.

Answer

$\text{A}=20\text{cm}^2=20\times10^{-4}\text{m}^2,$

$\text{T}=57^\circ\text{C}=330\text{K}$

$\text{E}=\text{A}\sigma\text{T}^4$

$=20\times10^{-4}\times6\times10^{-8}\times(330)^4\times10^4$

$=1.42\text{J}$

$\frac{\text{E}}{\text{t}}=\text{A}\sigma\text{e}\big({\text{T}_1^4}{}-\text{T}_2^4\big),$

$\text{A}=20\text{cm}^2=20\times10^{-4}\text{m}^2$

$\sigma=6\times10^{-8},$

$\text{T}_1=473\text{K},\ \text{T}_2=330\text{K}$

$=20\times10^{-4}\times6\times10^{-8}\times1\big[(473)^4-(330)^4\big]$

$=20\times6\times\big[5.005\times10^{-10}-1.185\times10^{10}\big]$

$=20\times6\times3.82\times10^{-2}$

$=4.58\text{w}$ From the ball.

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Question 275 Marks

A composite slab is prepared by pasting two plates of thicknesses L₁ and L₂ and thermal conductivities K₁ and K₂. The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

Answer

Here the thermal conductivities are in series,

$\therefore\frac{\frac{\text{K}_1\text{A}(\theta_1-\theta_2)}{\text{l}_1}\times\frac{\text{K}_2\text{A}(\theta_1-\theta_2)}{\text{l}_2}}{\frac{\text{K}_1\text{A}(\theta_1-\theta_2)}{\text{l}_1}+\frac{\text{K}_2\text{A}(\theta_1-\theta_2)}{\text{l}_2}}=\frac{\text{KA}(\theta_1-\theta_2)}{\text{l}_1+\text{l}_2}$

$\Rightarrow\frac{\frac{\text{K}_1}{\text{l}_1}\times\frac{\text{K}_2}{\text{l}_2}}{\frac{\text{K}_1}{\text{l}_1}+\frac{\text{K}_2}{\text{l}_2}}=\frac{\text{K}}{\text{l}_1+\text{l}_2}$

$\Rightarrow\frac{\text{K}_1+\text{K}_2}{\text{K}_1\text{l}_2+\text{K}_2\text{l}_1}=\frac{\text{K}}{\text{l}_1+\text{l}_2}$

$\Rightarrow\text{K}=\frac{(\text{K}_1\text{K}_2)(\text{l}_1+\text{l}_2)}{\text{K}_1\text{l}_2+\text{K}_2\text{l}_1}$

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Question 285 Marks

A room has a window fitted with a single 1.0m × 2.0m glass of thickness 2mm.

Calculate the rate of heat flow through the closed window when the temperature inside the room is 32^°C and that outside is 40^°C.
The glass is now replaced by two glasspanes, each having a thickness of 1mm and separated by a distance of 1mm. Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0Js^-1m^-1°C^-1 and that of air = 0.025Js^-1m^-1°C^-1.

Answer

$\frac{\text{Q}}{\text{t}}=\frac{\text{KA}(\theta_1-\theta_2)}{\ell}$

$=\frac{1\times2\times1(40-32)}{2\times10^{-3}}$

$=8000\text{J/sec}.$

Resistance of glass $=\frac{\ell}{\text{ak}_\text{g}}+\frac{\ell}{\text{ak}_\text{g}}$

Resistance of air $=\frac{\ell}{\text{ak}_\text{a}}$

Net resistance $=\frac{\ell}{\text{ak}_\text{g}}+\frac{\ell}{\text{ak}_\text{g}}+\frac{\ell}{\text{ak}_\text{a}}$

$=\frac{\ell}{\text{a}}\Big(\frac{2}{\text{k}_\text{g}}+\frac{1}{\text{k}_\text{a}}\Big)$

$=\frac{\ell}{\text{a}}\Big(\frac{2\text{k}_\text{a}+\text{k}_\text{g}}{\text{k}_\text{g}\text{k}_\text{a}}\Big)$

$=\frac{1\times10^{-3}}{2}\Big(\frac{2\times0.025+1}{0.025}\Big)$

$=\frac{1\times10^{-3}\times1.05}{0.05}$

$\frac{\text{Q}}{\text{t}}=\frac{\theta_1-\theta_2}{\text{R}}$

$=\frac{8\times0.05}{1\times10^{-3}\times1.05}$

$=380.9\approx381\text{W}$

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Question 295 Marks

A spherical ball A of surface area 20cm² is kept at the centre of a hollow spherical shell B of area 80cm². The surface of A and the inner surface of B emit as blackbodies. Both A and B are at 300K.

How much is the radiation energy emitted per second by the ball A?
How much is the radiation energy emitted per second by the inner surface of B
How much of the energy emitted by the inner surface of B falls back on this surface itself?

Answer

$\text{A}_\text{A}=20\text{cm}^2,\ \text{A}_\text{B}=80\text{cm}^2$

$(\text{mS})_\text{A}=40\text{J}^\circ\text{C},\ \text{(mS)}_\text{B}=82\text{J}^\circ\text{C},$

$\text{T}_\text{A}=100^\circ\text{C},\ \text{T}_\text{B}=20^\circ\text{C}$

K_B is low thus it is a poor conducter and K_A is high.

Thus A will absorb no heat and conduct all

$\Big(\frac{\text{E}}{\text{t}}\Big)_\text{A}=\sigma\text{A}_\text{A}\big[(373)^4-(293)^4\big]$

$\Rightarrow(\text{mS})_\text{A}\Big(\frac{\text{d}\theta}{\text{dt}}\Big)=\sigma\text{A}_\text{A}\big[(373)^4-(293)^4\big]$

$\Rightarrow\Big(\frac{\text{d}\theta}{\text{dt}}\Big)_\text{A}=\frac{\sigma\text{A}_\text{a}\big[(373)^4-(293)^4\big]}{(\text{mS})_\text{A}}$

$=\frac{6\times10^{-8}\big[(373)^4(293)^4\big]}{42}$

$=0.03^\circ\text{C/S}$

Similarly $\Big(\frac{\text{d}\theta}{\text{dt}}\Big)_\text{B}$

$=0.043^\circ\text{C/S}$

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Question 305 Marks

A hole of radius r₁ is made centrally in a uniform circular disc of thickness d and radius r₂. The inner surface (a cylinder of length d and radius r₁) is maintained at a temperature $\theta_1$ and the outer surface (a cylinder of length d and radius r₂) is maintained at a temperature $\theta_2(\theta_1>\theta_2).$ The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc.

Answer

$\frac{\text{dQ}}{\text{dt}}$ = Rate of flow of heat

Let us consider a strip at a distance r from the center of thickness dr.

$\frac{\text{dQ}}{\text{dt}}=\frac{\text{K}\times2\pi\text{r}\text{d}\times\text{d}\theta}{\text{dr}}$ $\big[\text{d}\theta=$ Temperature diff across the thickness dr$\big]$

$\Rightarrow\text{C}=\frac{\text{K}\times2\pi\text{r}\text{d}\times\text{d}\theta}{\text{dr}}$ $\Big[\text{C}=\frac{\text{d}\theta}{\text{dr}}\Big]$

$\Rightarrow\text{C}\int\limits^{\text{r}_2}_{\text{r}_1}\frac{\text{dr}}{\text{r}}=\text{K}2\pi\text{d}\int\limits_{\theta_1}^{\theta_2}\text{d}\theta$

$\Rightarrow\text{C}\Big[\log\Big]^{\text{r}_2}_{\text{r}_1}=\text{K}2\pi\text{d}(\theta_2-\theta_1)$

$\Rightarrow\text{C}(\log\text{r}_2-\log\text{r}_1)=\text{K}2\pi\text{d}(\theta_2-\theta_1)$

$\Rightarrow\text{C}\log\Big(\frac{\text{r}_2}{\text{r}_1}\Big)=\text{K}2\pi\text{d}(\theta_2-\theta_1)$

$\Rightarrow\text{C}=\frac{\text{K}2\pi\text{d}(\theta_2-\theta_1)}{\log\Big(\frac{\text{r}_2}{\text{r}_1}\Big)}$

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Question 315 Marks

Three rods of lengths 20cm each and area of cross section 1cm² are joined to form a triangle ABC. The conductivities of the rods are $\text{K}_\text{AB}=50\text{Js}^{-1}\text{m}^{-1}{^{\circ}}\text{C}^{-1},\ \text{K}_\text{BC}=200\text{Js}^{-1}\text{m}^{-1}{^{\circ}}\text{C}^{-1}$ and $\text{K}_\text{AC}=400\text{Js}^{-1}\text{m}^{-1}{^{\circ}\text{C}}^{-1}.$ The junctions A, B and C are maintained at 40^°C, 80^°C and 80^°C respectively. Find the rate of heat flowing through the rods AB, AC and BC.

Answer

$\text{K}_{\text{AB}}=50\text{j/m-s-}^\circ\text{C},\ \theta_\text{A}=40^\circ\text{C}$

$\text{K}_\text{BC}=200\text{j/m-s-}^\circ\text{C},\ \theta_\text{B}=80^\circ\text{C}$

$\text{K}_\text{AC}=400\text{j/m-s-}^\circ\text{C},\ \theta_\text{C}=80^\circ\text{C}$

Length $=20\text{cm}=20\times10^{-2}\text{m}$

$\text{A}=1\text{cm}^2=1\times10^{-4}\text{m}^2$

$\frac{\text{Q}_\text{AB}}{\text{t}}=\frac{\text{K}_\text{AB}\times\text{A}(\theta_\text{B}-\theta_\text{A})}{\text{l}}=\frac{50\times1\times10^{-4}\times40}{20\times10^{-2}}=1\text{W}.$
$\frac{\text{Q}_\text{AC}}{\text{t}}=\frac{\text{K}_\text{AC}\times\text{A}(\theta_\text{C}-\theta_\text{A})}{\text{l}}=\frac{400\times1\times10^{-4}\times40}{20\times10^{-2}}=800\times10^{-2}=8$
$\frac{\text{Q}_\text{BC}}{\text{t}}=\frac{\text{K}_\text{BC}\times\text{A}(\theta_\text{B}-\theta_\text{C})}{\text{l}}=\frac{200\times1\times10^{-4}\times0}{20\times10^{-2}}=0$

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Question 325 Marks

A cubical block of mass 1.0kg and edge 5.0cm is heated to 227^°C. It is kept in an evacuated chamber maintained at 27^°C. Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is 400Jkg^-1K^-1.

Answer

Since the Cube can be assumed as black body

$\text{e}=\ell$

$\sigma=6\times10^{-8}\text{w/m}^2\text{-k}^4$

$\text{A}=6\times25\times10^{-4}\text{m}^2$

$\text{m}=1\text{kg}$

$\text{s}=400\text{J/kg-}^\circ\text{K}$

$\text{T}_1=227^\circ\text{C}=500\text{K}$

$\text{T}_2=27^\circ\text{C}=300\text{K}$

$\Rightarrow\text{ms}\frac{\text{d}\theta}{\text{dt}}=\text{e}\sigma\text{A}\big(\text{T}_1^4-\text{T}_2^4\big)$

$\Rightarrow\frac{\text{d}\theta}{\text{dt}}=\frac{\text{e}\sigma\text{A}\big(\text{T}_1^4-\text{T}_2^4\big)}{\text{ms}}$

$=\frac{1\times6\times10^{-8}\times6\times25\times10^{-4}\times\big[(500)^4-(300)^4\big]}{1\times400}$

$=\frac{36\times25\times544}{}400\times10^{-4}$

$=0.1224^\circ\text{C/s}\approx0.12^\circ\text{C/s}.$

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Question 335 Marks

Seven rods A, B, C, D, E, F and G are joined as shown in figure. All the rods have equal cross-sectional area A and length l. The thermal conductivities of the rods are K_A = K_c = K₀, K_B = K_D = 2K₀, K_E= 3K_0, K_F = 4K₀, and K_G = 5K₀. The rod E is kept at a constant temperature T₂ and the rod G is kept at a constant temperature T₂(T₂ > T₁).

Show that the rod F has a uniform temperature $\text{T}=\frac{(\text{T}_1+2\text{T}_2)}{3}.$
Find the rate of heat flowing from the source which maintains the temperature T₂.

Answer

The temp at the both ends of bar F is same.
Rate of Heat flow to right = Rate of heat flow through left,
$\Rightarrow\Big(\frac{\text{Q}}{\text{t}}\Big)_\text{A}+\Big(\frac{\text{Q}}{\text{t}}\Big)_\text{C}=\Big(\frac{\text{Q}}{\text{t}}\Big)_\text{B}+\Big(\frac{\text{Q}}{\text{t}}\Big)_\text{D}$
$\Rightarrow\frac{\text{K}_\text{A}(\text{T}_1-\text{T})\text{A}}{\text{l}}+\frac{\text{K}_\text{C}(\text{T}_1-\text{T})\text{A}}{\text{l}}=\frac{\text{K}_\text{B}(\text{T}-\text{T}_2)\text{A}}{\text{l}}+\frac{\text{K}_\text{D}(\text{T}-\text{T}_2)\text{A}}{\text{l}}$
$\Rightarrow2\text{K}_0(\text{T}_1-\text{T})=2\times2\text{K}_0(\text{T}-\text{T}_2)$
$\Rightarrow\text{T}_1-\text{T}=2\text{T}-2\text{T}_2$
$\Rightarrow\text{T}=\frac{\text{T}_1+2\text{T}_2}{3}$

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