Mass of the metal, $m=0.20 kg =200 g$

Initial temperature of the metal, $T_{1}=150^{\circ} C$

Final temperature of the metal, $T_{2}=40^{\circ} C$

Calorimeter has water equivalent of mass, $m^{\prime}=0.025 kg =25 g$

Volume of water, $V=150 cm ^{3}$

Mass $(M)$ of water at temperature $T=27^{\circ} C$

$150 \times 1=150 g$

Fall in the temperature of the metal

$\Delta T=T_{1}-T_{2}=150-40=110^{\circ} C$

Specific heat of water, $C_{ w }=4.186 J / g /^{\circ} K$

Specific heat of the metal $=C$

Heat lost by the metal, $\theta=m C \Delta T$

Rise in the temperature of the water and calorimeter system:

$\Delta T^{\prime}=40-27=13^{\circ} C$

Heat gained by the water and calorimeter system:

$\Delta \theta^{\prime \prime}=m_{1} C_{ w } \Delta T^{\prime}$

$=\left(M+m^{\prime}\right) C_{ w } \Delta T^{\prime}$

Heat lost by the metal $=$ Heat gained by the water and colorimeter system

$m C \Delta T=(M+m) C_{ w } \Delta T$

$200 \times C \times 110=(150+25) \times 4.186 \times 13$

$\therefore C=\frac{175 \times 4.186 \times 13}{110 \times 200}=0.43 Jg ^{-1} K ^{-1}$

If some heat is lost to the surroundings, then the value of $C$ will be smaller than the actual value.