Question

An unknown metal of mass 0.280 kg is heated to 160.0°C and dropped in an aluminum calorimeter of mass 0.250 kg that contains 0.170 kg of water at 30°C. The calorimeter, water, and unknown metal have a final temperature of 46.0°C. Find the specific heat of the unknown metal. Hint: you need the specific heat of water and aluminum. Use units of [J/(kg.K)] and the values in your book for the specific heat.

Answer 1

Using principle of caloriemetry

Heat gained by the caloriemeter + water = Heat lost by the metal

specific heat capacity of aluminium is S_al = 900 J/Kg-K

specific heat capacity of water is S_w = 4186 J/Kg-K

then

Heat gained by the caloriemeter + water = Heat lost by the metal

Q1+Q2 = Q3

(m_Al*S_Al*dT1) + (m_w*S_w*dT2) = (m_m*S_m*dT3)

(0.25*900*(46-30))+(0.17*4186*(46-30)) = (0.28*S_m*(160-46))

S_m = 469.5 J/Kg-K is the specific heat capacity of unknown metal