Answer In solving this example, we shall use the fact that at a steady state, heat given by an aluminium sphere will be equal to the heat absorbed by the water and calorimeter.

Mass of aluminium sphere $\left(m_{1}\right)=0.047 kg$

Initial temperature ofaluminiumsphere $=100^{\circ} C$

Final temperature $=23^{\circ} C$

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

Let specific heat capacity of aluminium be $s_{ Al }$.

The amount of heat lost by the aluminium sphere $=m_{1} s_{A l} \Delta T=0.047 kg \times s_{A l} \times 77^{\circ} C$

Mass of water $\left(m_{2}\right)=0.25 kg$

Mass of calorimeter $\left(m_{3}\right)=0.14 kg$

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

Final temperature of the mixture $=23^{\circ} C$

Change in temperature $\left(\Delta T_{2}\right)=23^{\circ} C -20^{\circ} C =3^{\circ} C$

Specific heat capacity of water $\left(s_{w}\right)$

$=4.18 \times 10^{3} J kg ^{-1} K ^{-1}$

Specific heat capacity of copper calorimeter $=0.386 \times 10^{3} J kg ^{-1} K ^{-1}$

The amount of heat gained by water and calorimeter $=m_{2} s_{w} \Delta T_{2}+m_{3} s_{ ca } \Delta T_{2}$

$=\left(m_{2} s_{w}+m_{3} s_{ cu }\right)\left(\Delta T_{2}\right)$

$=\left(0.25 kg \times 4.18 \times 10^{3} J kg ^{-1} K ^{-1}+0.14 kg \times\right.$

$\left.0.386 \times 10^{3} J kg ^{-1} K ^{-1}\right)\left(23^{\circ} C -20^{\circ} C \right)$

In the steady state heat lost by the aluminium sphere $=$ heat gained by water + heat gained by calorimeter. So, $0.047 kg \times s_{ A 1} \times 77^{\circ} C$

$=\left(0.25 kg \times 4.18 \times 10^{3} J kg ^{-1} K ^{-1}+0.14 kg \times\right.$

$\left.0.386 \times 10^{3} J kg ^{-1} K ^{-1}\right)\left(3^{\circ} C \right)$

$s_{A t}=0.911 kJ kg ^{-1} K ^{-1}$