Question

Suppose the masses of the calorimeter, the water in it and the hot object made up of copper which is put in the calorimeter are the same. The initial temperature of the calorimeter and water is ${30}^{\circ }C$ and that of the hot object is ${60}^{\circ }C$. The specific heats of copper and water are $0.09cal{g}^{-1}{}^{\circ }{C}^{-1}$ and $1cal{g}^{-1}{}^{\circ }{C}^{-1}$ respectively. What will be the final temperature of water?

• A. ${21.54}^{\circ }C$
• B. ${15.53}^{\circ }C$
• C. ${32.29}^{\circ }C$
• D. ${46.21}^{\circ }C$

Answer 1

The correct option is C ${32.29}^{\circ }C$

Let the mass of the water is $m_w$, mass of the hot object is $m_o$ and mass of the calorimeter is $m_c$.
Initial temperature of the calorimeter and water is $T_i$, temperature of hot object is $T_o$ and final temperature of water is $T_f$.
Given: $m_w=m_o=m_c=m$, $T_i={30}^{\circ }C$, $T_o={60}^{\circ }C$, $T_f=?$
Using the relation,
$m_o\times \Delta T_o\times c_o=m_w\times \Delta T_w\times c_w+m_c\times \Delta T_c\times c_c$
where, $\Delta T_o$ is the change in temperature of hot object, $c_o$ is the specific heat capacity of hot object, $\Delta T_w$ is the change in temperature of water and $c_w$ is the specific heat of water.
Given, $c_o=0.09cal{g}^{-1}{}^{\circ }{C}^{-1}$ and $c_w=1cal{g}^{-1}{}^{\circ }{C}^{-1}$
or, $m\times (60-T_f)\times 0.09=m\times (T_f-30)\times 1+m\times (T_f-30)\times 0.09$
or, $(60-T_f)\times 0.09=(T_f-30)\times 1.09$
or, $60\times 0.09+30\times 1.09=(0.09+1.09)T_f$
$T_f={32.29}^{\circ }C$
The final temperature of water will be ${32.29}^{\circ }C$.