In an experiment on the specific heat of a metal,a $0.20 \; kg$ block of the metal at $150 \; ^{\circ}C$ is dropped in a copper calorimeter (of water equivalent $0.025 \; kg$) containing $150 \; cm^{3}$ of water at $27 \; ^{\circ}C$. The final temperature is $40 \; ^{\circ}C$. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible,is your answer greater or smaller than the actual value for the specific heat of the metal?

(0.433 J/GK) 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,$T_{2} = 40 \; ^{\circ}C$.
Water equivalent of the calorimeter,$m' = 0.025 \; kg = 25 \; g$.
Volume of water,$V = 150 \; cm^{3}$.
Mass of water,$M = 150 \; g$ (since density is $1 \; g/cm^{3}$).
Fall in temperature of the metal,$\Delta T = 150 - 40 = 110 \; ^{\circ}C$.
Rise in temperature of the water and calorimeter,$\Delta T' = 40 - 27 = 13 \; ^{\circ}C$.
Specific heat of water,$C_{w} = 4.186 \; J \cdot g^{-1} \cdot K^{-1}$.
Heat lost by metal = Heat gained by water and calorimeter.
$m \cdot C \cdot \Delta T = (M + m') \cdot C_{w} \cdot \Delta T'$.
$200 \cdot C \cdot 110 = (150 + 25) \cdot 4.186 \cdot 13$.
$22000 \cdot C = 175 \cdot 4.186 \cdot 13 = 9523.15$.
$C = 9523.15 / 22000 \approx 0.433 \; J \cdot g^{-1} \cdot K^{-1}$.
If heat is lost to the surroundings,the measured heat gained by the water/calorimeter is less than the actual heat lost by the metal. Thus,the calculated specific heat $C$ would be smaller than the actual value.