Question

An unknown object is initially at a temperature of 300 degrees and has a mass of .3Kg. It is placed in 1 Kg of water initially at a temperature of 10 degrees. If the final temperature is 13degrees water is the specific heat of the object?

Answer 1

Suppose the specific heat of the object = s

Mass of unknown object, m1 = 0.3 kg

Initial temperature, T1 = 300 deg C

Mass of water, m2 = 1.0 kg

Initial temperature of water, T2 = 10 deg C

Final temperature of unknown object and water, T = 10 deg C.

Specific heat capacity of water = 4200 J/kg*C

Heat loss by the unknown object, Q1 = m1*(T1 - T)*s

Heat gained by water, Q2 = m2*(T - T2)*4200

Now,

Heat loss = Heat gain

=> Q1 = Q2

=> m1*(T1 - T)*s = m2*(T - T2)*4200

=> 0.3*(300 - 13)*s = 1.0*(13 - 10)*4200

=> s = [1.0*(13 - 10)*4200] / [0.3*(300 - 13)]

= [3*4200] / [0.3*287] = 146.3 J/kg*C

So, specific heat capacity of the unknown object = s = 146.3 J/kg*C