Question

Calculate the specific heat of a metal (in calories/gram-degree C) from the following data. A container made of the metal has a mass of 3.80 kg and contains 13.1 kg of water. A 1.40 kg piece of the same metal, initially at a temperature of 140 degrees C, is placed in the water. The container and water initially have a temperature of 15 degrees C, and the final temperature of the entire system is 18 degrees C.

Answer 1

A balance equation (the balance of thermal energy) is

Cm = 3.8*(18-15)+Cw*(13.1*(18-15) = Cm = 1.4*(140-18)

It says "amount of heat absorbed by the system to raise its temperature from 15°C to 18°C is the same as the
amount of heat
lost by the metal piece when it cooled from 140°C to 15°C).

Hear Cm is the specific heat of the metal ( in calories/(gram-degree C) ), which is under the question),

Cw is the specific heat of water ( = 1 calorie/(gram-degree C) ).

From (1), express

Cm = 13.1*(18-15)/1.4*(140-18)-3.8*(18-15) = 0.2266calorie/(gram-degree C)