Question

A calorimeter contains $0.2kg$ of water at $30^\circ C.$ $0.1kg$ of water at $60^\circ C$is added to it, the mixture is well stirred and the resulting temperature is found to be $35^\circ C$. The thermal capacity of the calorimeter is
A. $6300J/K$
B. $1260J/K$
C. $4200J/K$
D. none of these

Answer 1

Calculate the total heat gained and total heat lost. Then by using the principle of calorimetry, solve the question.

Complete step by step answer: It is given in the question that,
The calorimeter contain ${m_1} = 0.2kg$ of water at ${t_1} = {30^0}C$
${m_2} = 0.1kg$ of water at ${t_2} = {60^0}C$
The resulting temperature is $t = {35^0}C$
The specific heat of water is $4200Jk{g^{ - 1}}K$
Let $C$ be the thermal capacity of water.
We know that the heat gained by a calorimeter is equal to the product of thermal capacity and the rise in temperature.
Since, the initial minimum temperature in calorimeter was ${30^0}C$ and the temperature of the mixture was found to be ${35^0}C$
From the above explanation, we get
${h_c} = C({35^0} - {30^0})$
$\Rightarrow {h_c} = 5C$
Where, ${h_c}$ is the heat gained by a calorimeter.
Now, the heat gained by water is the product of mass of water for which the heat was gained, specific heat and the rise in temperature.
Since, the mass of water that gained heat was ${m_1} = 0.2kg$ at the temperature of ${30^0}C$ and it rose to ${35^0}C$ in the mixture.
Therefore, using above given explanation, we can write
${h_{w1}} = 0.2 \times 4200 \times ({35^0} - {30^0})$
$\Rightarrow {h_{w1}} = 0.2 \times 4200 \times {5^0}$
$\Rightarrow {h_{w1}} = 4200$
Where, ${h_{w1}}$ is the heat gained by water of mass ${m_1}$
Therefore, total heat gained will be the sum of heat gained by the calorimeter and heat gained by the water.
$\Rightarrow {h_{tg}} = 5C + 4200$
Where,
${h_{tg}}$ is total heat gained
Now, heat lost by $0.1kg$ of hot water, when added to calorimeter is the product of the mass of water, specific heat and the loss in temperature.
$\Rightarrow {h_{w2}} = 0.1 \times 4200 \times ({60^0} - {35^0})$
$\Rightarrow {h_{w2}} = 420 \times 25$
$\Rightarrow {h_{w2}} = 10500$
According to the principle of calorimetry,
Heat gained is equal to heat lost.
$\Rightarrow 5C + 4200 = 10500$
$\Rightarrow 5C = 10500 - 4200$
$\Rightarrow 5C = 6300$
$\Rightarrow C = 1260J/K$

Therefore the correct answer is (B).

Note: Since heat is a scalar quantity, we do not use negative signs to show heat loss. But we write that it was a heat loss even though the value is positive. Important part of this question was the understanding of the principle of calorimetry.