5 Marks Questions

Question 15 Marks

A piece of iron of mass $100g$ is kept inside a furnace for a long time and then put in a calorimeter of water equivalent $10g$ containing $240g$ of water at $20°C$. The mixture attains and equilibrium temperature of $60°C$. Find the temperature of the furnace. Specific heat capacity of iron = $470J kg^{-1 \circ} C^{-1}$.

Answer

Given, Mass of iron = 100g Water equivalent of calorimeter = 10g Mass of water = 240gm Let the temperature of surface be $\theta^\circ\text{C}.$ Specific heat capacity of iron = $470J kg^{-1 \circ} C^{-1}$ Total heat gained = Total heat lost$\Rightarrow\frac{100}{1000}\times470\times(\theta-60^\circ)$
$=\frac{(240+10)}{1000}\times4200\times(60-20)$
$\Rightarrow47\theta-47\times60=25\times42\times40$
$\Rightarrow\theta=\frac{42000+2820}{47}=\frac{44820}{47}$
$\Rightarrow953.61^\circ\text{C}$

View full question & answer→

Question 25 Marks

The temperatures of equal masses of three different liquids $A , B$ and C are $12^{\circ} C , 19^{\circ} C$ and $28^{\circ} C$ respectively. The temperature when A and B are mixed is $16^{\circ} C$, and when B and C are mixed, it is $23^{\circ} C$. What will be the temperature when A and C are mixed?

Answer

Given, Temperature of $A =12^{\circ} C$
Temperature of $B =19^{\circ} C$
Temperature of $C =28^{\circ} C$
Temperature of mixture of $A$ and $B=16^{\circ} C$
Temperature of mixture of B and $C =23^{\circ} C$
Let the mass of the mixtures be M and the specific heat capacities of the liquids A, B and C be $C_A, C_B, $and $C_C,$ respectively. According to the principle of calorimetry, when A and B are mixed,
we get Heat gained by Liquid A = Heat lost by liquid B
$\Rightarrow MC_A(16 - 12) = MC_B(19 - 16)$
$\Rightarrow 4MC_A =$ 3MCB$\Rightarrow\text{MC}_\text{A}=\Big(\frac{3}{4}\Big)\text{MC}_\text{B}\ \dots(1)$
When B and C are mixed, Heat gained by liquid B = Heat lost by liquid C
$\Rightarrow MC_B(23 - 19) = MC_C(28 - 23)$
$\Rightarrow 4MC_B =$ 5MCC$\Rightarrow\text{MC}_\text{C}=\Big(\frac{4}{5}\Big)\text{MC}_\text{B}\ \dots(2)$
When A and C are mixed, Let the temperature of the mixture be T.
Then, Heat gained by liquid A = Heat lost by liquid C
$\Rightarrow MC_A (T - 12) = MC_C (28 - T)$
Using the values of $MC_A$and $MC_C,$
we get From eqs. (1) and (2),
$\Rightarrow\Big(\frac{3}{4}\Big)\text{MC}_\text{B}(\text{T}-12)=\Big(\frac{4}{5}\Big)\text{MC}_\text{B}(28-\text{T})$
$\Rightarrow\Big(\frac{3}{4}\Big)(\text{T}-12)=\Big(\frac{4}{5}\Big)(28-\text{T})$
$\Rightarrow(3\times5)(\text{T}-12)=(4\times4)(28-\text{T})$
$\Rightarrow15\text{T}-180=448-16\text{T}$
$\Rightarrow31\text{T}=628$
$\Rightarrow\text{T}=\frac{628}{31}=20.253^\circ\text{C}$
$\Rightarrow\text{T}=20.3^\circ\text{C}$

View full question & answer→

Question 35 Marks

An aluminium vessel of mass 0.5kg contains 0.2kg of water at $20^\circ C$. A block of iron of mass 0.2kg at $100^\circ C$ is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are $910\ J\ kg^{- 1}-K^{- 1}, 470\ J\ kg^{-1}-K^{-1} $ and $4200\ J\ kg^{-1}-K^{-1} $ respectively.

Answer

Given, Mass of aluminium $= 0.5kg$
Mass of water $= 0.2kg$
Mass of iron $= 0.2kg$
Specific heat of aluminium $= 910\ J\ kg^- 1-K^{-1}$
Specific heat of iron $= 470\ J\ kg^{- 1}\ K^{- 1}$
Specific heat of water $= 4200J\ kg^{- 1}\ K^{- 1}$
Let the equilibrium temperature of the mixture be $T.$
Temperature of aluminium and water $= 20^\circ C = 273 + 20 = 293K$
Temperature of iron $= 100^\circ C = 273 + 100 = 373K$
Heat lost by iron, $H_1 = 0.2 \times 470 \times (373 - T)$
Heat gained by water $= 0.2 \times 4200 \times (T - 293)$
Heat gained by iron $= 0.5 \times 910 \times (T - 293)$
Total heat gained by water and iron,
$H_2 = 0.5 \times 910 (T - 293) + 0.2 \times 4200 \times (T - 293) $
$H_2 = (T - 293) [0.5 \times 910 + 0.2 \times 4200]$
We know, Heat gain = Heat lost
$\Rightarrow (T - 293)[0.5 \times 910 + 0.2 \times 4200] = 0.2 \times 470 \times (373 - T)$
$ \Rightarrow (T - 293)(455 + 840) = 94(373 - T)$
$\Rightarrow (\text{T} - 293)\frac{1295}{94}=(373−\text{T}) $
$\Rightarrow (T - 293) \times 14 = (373 - T) $
$\Rightarrow 14T - 293 \times 14 = 373 - T $
$\Rightarrow 15T = 373 + 4102 = 4475$
$\Rightarrow\text{T}=\frac{4475}{15}=298.33\text{K}\approx298\text{K}$
$\therefore T = (298 - 273)^\circ C = 25^\circ C$
$\therefore$ Final temperature $= 25^\circ C$

View full question & answer→

Physics 5 Marks Questions

Take a timed test