Atemperature 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.

Explanation:

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

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