Question

A calorimeter of mass $50g$ and specific heat capacity $0.42\,J\,{g^{ - 1o}}{C^{ - 1}}$ contains some mass of water at ${20^o}C$. A metal piece of mass $20\,g$ at ${100^o}C$ is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be ${22^o}C$. Find the mass of water used in the calorimeter.
[Specific heat capacity of the metal piece $ = 0.3\,J\,{g^{ - 1}}^o{C^{ - 1}}$
Specific heat capacity of water$ = 4.2\,J\,{g^{ - 1}}^o{C^{ - 1}}$]

Answer 1

Hint: While solving the question we must to keep in mind the concept of calorimetry. The principle of calorimetry indicates the law of conservation energy, i.e. the total heat lost by the hot body is equal to the total heat gained by the cold body.

Complete step by step answer:
Heat energy given by metal piece
$ = m.c.\Delta {T_1}$
$ = 20 \times 0.3 \times (100 - 22)$
$ = 468\,J$
Heat energy gained by water
${m_w} \times {c_w} \times \Delta {T_2}$
$ = {m_w} \times 4.2 \times (22 - 20)$
$ = {m_w} \times 8.4\,J$
Heat energy gained by calorimeter
$ = {m_C} \times {c_c} \times \Delta {T_2}$
$ = 50 \times 0.42 \times (22 - 20)$
$ = 42\,Joule.$
By principle of calorimeter
Heat load $ = $heat gained
Heat energy given by metal
$ = $Heat energy gained
By water$ + $Heat energy gained by calorimeter.
$468 = \left( {{m_w} \times 8.4} \right) + 42$
${m_w} = 50.7\,g$
Therefore mass $ + $water used in calorimeter is $50.7\,g$

Note:
When two bodies of different temperatures (preferably a solid and a liquid) are placed in physical contact with each other, the heat is transferred from the body with higher temperature to the body with lower temperature until thermal equilibrium is attained between them.