Question

Two identical calorimeters A and B contain equal quantities of water at ${20^{\text{o}}}{\text{C}}$. A ${\text{5}}\,{\text{gm}}$ piece of metal X of specific heat ${\text{0}}{\text{.2 cal }}{{\text{g}}^{ - 1}}{\left( {^{\text{o}}{\text{C}}} \right)^{ - 1}}$ is dropped into A and a ${\text{5}}\,{\text{gm}}$ piece of metal Y into B. The equilibrium temperature in A is ${22^{\text{o}}}{\text{C}}$ and in B ${23^{\text{o}}}{\text{C}}$. The initial temperature of both the metals is ${40^{\text{o}}}{\text{C}}$. Find the specific heat of metal Y in ${\text{cal }}{{\text{g}}^{ - 1}}{\left( {^{\text{o}}{\text{C}}} \right)^{ - 1}}$.

Answer 1

We are asked to find the specific heat capacity of metal Y. Here, you will need to use the formula for heat transfer and also you have to apply the principle of calorimetry. Apply the principle of calorimetry for each calorimeter A and B separately and using the equations obtained, find the value of specific heat of metal Y.

Complete step by step answer:
Given, temperature of water in calorimeter A, ${T_A} = {20^{\text{o}}}{\text{C}}$
Temperature of water in calorimeter B, ${T_B} = {20^{\text{o}}}{\text{C}}$
Mass of metal X, ${m_X} = {\text{5}}\,{\text{gm}}$
Mass of metal Y, ${m_Y} = {\text{5}}\,{\text{gm}}$
Specific heat capacity of metal X, ${s_X} = {\text{0}}{\text{.2 cal }}{{\text{g}}^{ - 1}}{\left( {^{\text{o}}{\text{C}}} \right)^{ - 1}}$
Temperature of metal X, ${T_X} = {40^{\text{o}}}{\text{C}}$
Temperature of metal Y, ${T_Y} = {40^{\text{o}}}{\text{C}}$
Equilibrium temperature of calorimeter A, ${T_{eA}} = {22^{\text{o}}}{\text{C}}$
Equilibrium temperature of calorimeter B, ${T_{eB}} = {23^{\text{o}}}{\text{C}}$
Let the specific heat capacity of metal Y be ${s_Y}$ and the mass of water of both the calorimeter A and B be ${m_W}$.

We will use here the formula of heat transfer which is,
$Q = ms\Delta T$ (i)
where $m$ is the mass of the substance, $s$ is the specific heat capacity and $\Delta T$ is the difference between the initial and equilibrium temperature.
This formula is used to calculate heat loss or gain by a substance.
The specific heat of water is ${s_W} = 1\,{\text{cal }}{{\text{g}}^{ - 1}}{\left( {^{\text{o}}{\text{C}}} \right)^{ - 1}}$
According to the principle of calorimetry we have, heat loss by the hot body is equal to heat gain by the cold body.
So, here the heat loss by metal X and Y will be equal to heat gain by calorimeter A and B respectively.

For calorimeter A,
${\text{Heat loss by metal X}} = {\text{Heat gain by water}}$ (ii)
Heat loss by metal X using the formula from equation (i) is,
${\text{Heat loss by metal X}} = {m_X}{s_X}\left( {{T_X} - {T_{eA}}} \right)$
Putting the values of ${m_X}$, ${s_X}$, ${T_X}$ and ${T_{eA}}$ we get,
${\text{Heat loss by metal X}} = 5 \times 0.2 \times \left( {40 - 22} \right)$
${\text{Heat loss by metal X}} = 18\,{\text{cal}}$
Heat gain by water using the formula from equation (i) is,
${\text{Heat gain by water}} = {m_W}{s_W}\left( {{T_{eA}} - {T_W}} \right)$
Putting the values of ${s_W}$, ${T_{eA}}$ and ${T_W}$ we get,
${\text{Heat gain by water}} = {m_W} \times 1 \times \left( {22 - 20} \right)$
${\text{Heat gain by water}} = 2{m_W}$
Now substituting the values of heat loss and heat gain in equation (ii), we get
$18\, = 2{m_W}$
$\therefore{m_W} = 9\,{\text{gm}}$

For calorimeter B,
${\text{Heat loss by metal Y}} = {\text{Heat gain by water}}$ (iii)
Heat loss by metal Y using the formula from equation (i) is,
${\text{Heat loss by metal Y}} = {m_Y}{s_Y}\left( {{T_Y} - {T_{eB}}} \right)$
Putting the values of ${m_Y}$, ${T_Y}$ and ${T_{eB}}$ we get,
${\text{Heat loss by metal Y}} = 5 \times {s_Y} \times \left( {40 - 23} \right)$
${\text{Heat loss by metal Y}} = 85{s_Y}$
Heat gain by water using the formula from equation (i) is,
${\text{Heat gain by water}} = {m_W}{s_W}\left( {{T_{eB}} - {T_W}} \right)$
Putting the values of ${m_W}$, ${s_W}$, ${T_{eB}}$ and ${T_W}$ we get,
${\text{Heat gain by water}} = 9 \times 1 \times \left( {23 - 20} \right)$
${\text{Heat gain by water}} = 27\,{\text{cal}}$
Now substituting the values of heat loss and heat gain in equation (iii), we get
$85{s_Y}\, = 27$
${s_Y} = \dfrac{{27}}{{85}}$
$\therefore{s_Y} = {\text{0}}{\text{.318 cal }}{{\text{g}}^{ - 1}}{\left( {^{\text{o}}{\text{C}}} \right)^{ - 1}}$

Therefore, specific heat of metal Y is ${\text{0}}{\text{.318 cal }}{{\text{g}}^{ - 1}}{\left( {^{\text{o}}{\text{C}}} \right)^{ - 1}}$.

Note: We have used the principle of calorimetry. If we try understand the principle of calorimetry in detail, it says that of two bodies which are at different temperatures are placed in contact with each other then the body at higher temperature releases heat energy to the body at lower temperature which absorbs the heat, which is why here we taken heat loss by the metal to be equal to heat gain of the water as, metal is at higher temperature than water.