5 Marks Questions

Question 15 Marks

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

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
$\Rightarrow\frac{100}{1000}\times470\times(\theta-60^\circ)$
$=\frac{(240+10)}{1000}\times4200\times(60-20)$
$\Rightarrow47\theta-47\times60=25\times42\times40$
$\Rightarrow\theta=\frac{42000+2820}{47}=\frac{44820}{47}$
$\Rightarrow953.61^\circ\text{C}$

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Question 25 Marks

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

Answer

Given,
Temperature of A = 12°C
Temperature of B = 19°C
Temperature of C = 28°C
Temperature of mixture of A and B = 16°C
Temperature of mixture of B and C = 23°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
$\Rightarrow MC_A(16 - 12) = MC_B(19 - 16)$
$\Rightarrow 4MC_A = 3MC_B$
$\Rightarrow\text{MC}_\text{A}=\Big(\frac{3}{4}\Big)\text{MC}_\text{B}\ \dots(1)$
When B and C are mixed,
Heat gained by liquid B = Heat lost by liquid C
$\Rightarrow MC_B(23 - 19) = MC_C(28 - 23)$
$\Rightarrow 4MC_B = 5MC_C$
$\Rightarrow\text{MC}_\text{C}=\Big(\frac{4}{5}\Big)\text{MC}_\text{B}\ \dots(2)$
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),
$\Rightarrow\Big(\frac{3}{4}\Big)\text{MC}_\text{B}(\text{T}-12)=\Big(\frac{4}{5}\Big)\text{MC}_\text{B}(28-\text{T})$
$\Rightarrow\Big(\frac{3}{4}\Big)(\text{T}-12)=\Big(\frac{4}{5}\Big)(28-\text{T})$
$\Rightarrow(3\times5)(\text{T}-12)=(4\times4)(28-\text{T})$
$\Rightarrow15\text{T}-180=448-16\text{T}$
$\Rightarrow31\text{T}=628$
$\Rightarrow\text{T}=\frac{628}{31}=20.253^\circ\text{C}$
$\Rightarrow\text{T}=20.3^\circ\text{C}$

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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 $910Jkg^{-1}-K^{-1}$, $470Jkg^{-1}-K^{-1}$ and $4200Jkg^{-1}-K^{-1}$ respectively.

Answer

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

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