A temperature of a 50 kg block increases by 15°C when 337,500 J of thermal energy are added to the block.

a. What is the specific heat of the object? show your work with units.

Heat = mass * specific heat * temperature change

q = m C ΔT

Given:

q = 337500 J

m = 50 kg

ΔT = 15°C

Substitute:

337500 J = (50 kg) C (15°C)

C = 450 J/kg/°C

Specific heat is usually recorded in J/g/°C or kJ/kg/°C. Converting:

C = 0.45 J/g/°C = 0.45 kJ/kg/°C