Question

Two identical calorimeter $A$ and $B$ contain equal quantity of water at ${20}^{\circ }C$. A $5gm$ piece of metal $X$ of specific heat $0.2cal{g}^{-1}(C^{\circ }{)}^{-1}$ is dropped into $A$ and a $5gm$ piece of metal $Y$ into $B$. The equilibrium temperature in $A$ is ${22}^{\circ }C$ and in $B{23}^{\circ }C$. The initial temperature of both the metals is ${40}^{\circ }C$. Find the specific heat of metal $Y$ in cal $g^{-1}(C^{\circ }{)}^{-1}$.

Answer 1

Given: Two identical calorimeter A and B contain equal quantity of water at ${20}^{\circ }C$. A $5gm$ piece of metal $X$ of specific heat $0.2cal{g}^{-1}{(}^{\circ }C{)}^{-1}$ is dropped into A and a $5gm$ piece of metal Y into B. The equilibrium temperature in A is ${22}^{\circ }C$ and in B ${23}^{\circ }C$. The initial temperature of both the metals is ${40}^{\circ }C$.

To find the specific heat of metal Y in $cal{g}^{-1}{(}^{\circ }C{)}^{-1}.$

Solution:

As per the given condition,

Mass of metal X, $m_x=5g$

Mass of metal Y, $m_y=5g$

Temperature difference of the water for metal X, $\Delta T_{w_x}=22-20=2^{\circ }C$

Temperature difference of the water for metal Y, $\Delta T_{w_y}=23-20=3^{\circ }C$

Temperature difference of metal X, $\Delta T_x=40-22={18}^{\circ }C$

Temperature difference of the metal Y, $\Delta T_y=40-23={17}^{\circ }C$

Specific heat of metal X, $s_x=0.2cal{g}^{-1}{(}^{\circ }C{)}^{-1}$

Specific heat of water, $s_w=1cal{g}^{-1}{(}^{\circ }C{)}^{-1}$

Let the mass of the water be $m_w=x$and specific heat of metal Y be $s_y$

We know in a calorimeter,

Heat lost by metal= heat gained by water

In calorimeter $A$, for metal $X$;

$m_x{s}_x\Delta T_x=m_w{s}_w\Delta T_{w_x}⟹5\times 0.2\times 18=x\times 1\times 2⟹x=9gm$

is the mass of the water, both calorimeter has same mass of water in it.

In calorimeter $B$, for metal $Y$;

$m_y{s}_y\Delta T_y=m_w{s}_w\Delta T_{w_y}⟹5\times s_y\times 17=x\times 1\times 3⟹s_y=\frac{9\times 3}{5\times 17}=0.318cal{g}^{-1}{(}^{\circ }C{)}^{-1}$

is the specific heat of metal Y