Question

An aluminium sphere of mass $0.047kg$ is heated to ${100}^{\circ }C$. It is dropped in a copper calorimeter of mass $0.14kg$, containing $0.25kg$ of water at ${20}^{o}C$. The temperature of water rises to a steady state at ${23}^{o}C$. Calculate the specific heat of aluminium. specific heat of water $=4.18\times {10}^{3}Jk{g}^{-1}{}^o{C}^{-1}$, specific heat of copper $0.386\times {10}^{3}Jk{g}^{-1}{}^o{C}^{-1}$.

• A. $191Jk{g}^{-1}{}^o{C}^{-1}$
• B. $911Jk{g}^{-1}{}^o{C}^{-1}$
• C. $119Jk{g}^{-1}{}^o{C}^{-1}$
• D. $0.911Jk{g}^{-1}{}^o{C}^{-1}$

Answer 1

The correct option is B $911Jk{g}^{-1}{}^o{C}^{-1}$

solutions:

Mass of the aluminium sphere $\left(m_1\right)=0.047kg$

Initial temperature of the aluminium sphere$={100}^{o}C$

Final temperature$={20}^{o}C$

Change in temperature $\left(\Delta T_1\right)={100}^{o}C-{23}^{o}C={77}^{o}C$

Let, specific heat capacity of $Al=c_1$

​ Amount of heat lost by $Al=m1\times c1=\Delta T_1$ ​ = $0.047kg\times c_1\times {77}^{o}c$ Mass of the water $m_2=0.25kg$

Mass of the calorimeter $m_3=0.14kg$

Initial temperature of water and calorimeter$={20}^{o}C$

Final temperature $={23}^{o}C$

Change in temperature $\left(\Delta T_2\right)={23}^{o}C-{20}^{o}C=3^{o}C$

Specific heat capacity of water $\left(c_2\right)=4.18\times {10}^{3}Jk{g}^{-1}{C}^{-1}$

Total amount of heat gained by calorimeter and water $=m_2\times c_2\times \Delta T_2+m_3\times c_3\times \Delta T_2$ ​ =$0.25\times 4.18\times {10}^3\times 3^{o}C+0.14\times 0.386\times {10}^3\times 3^{o}C=3297.12J$

At steady state, heat lost by Al= heat gained by water+ heat gained by calorimeter

Therefore $0.047kg\times c_1\times {77}^{o}c=3297.12J$

$c_1=911.058Jk{g}^{-1}{c}^{-1}$