5 Marks Questions

Question 15 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
$\Rightarrow MC_A(16 - 12) = MC_B(19 - 16)$
$\Rightarrow 4MC_A = 3MC_B$
$\Rightarrow\text{MC}_\text{A}=\Big(\frac{3}{4}\Big)\text{MC}_\text{B}\ \dots(1)$
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 25 Marks

An aluminium vessel of mass $0.5\ kg$ contains $0.2\ kg$ of water at $20^\circ C$. A block of iron of mass $0.2\ kg$ 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 $= 0.5\ kg$
Mass of water $= 0.2\ kg$
Mass of iron $= 0.2\ kg$
Specific heat of aluminium $= 910\ J\ kg^{-1}-K^{-1}$
Specific heat of iron $= 470\ J\ kg^{-1} K^{-1}$
Specific heat of water $= 4200\ J \ kg^{-1 }K^{-1}$
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 $= 25^\circ C$

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Question 35 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 $20^\circ C.$ The mixture attains and equilibrium temperature of $60^\circ C$. 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
$\Rightarrow\frac{100}{1000}\times470\times(\theta-60^\circ)$
$=\frac{(240+10)}{1000}\times4200\times(60-20)$
$\Rightarrow47\theta-47\times60=25\times42\times40$
$\Rightarrow\theta=\frac{42000+2820}{47}=\frac{44820}{47}$
$\Rightarrow953.61^\circ\text{C}$

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