Question

A 0.200 kg block of a substance requires 3.59 kJ of heat to raise its temperature by 20.0 K. What is the specific heat of the substance? A. 2,020 J/(kg * K) B. 383 J/(kg*K) C. 130 J/(kg * K) D. 897 J/(kg*K)

Answer 1

Answer is D. The formula is Q=c*m*delta T. To raise m kg of substance with specific heat c by T kelvin, we need Q J of heat. Rearrange the equation, c=Q/(m*delta T). Plug in our numbers, Q=3.59kJ=3590J, m=0.200, T=20.0, c=3590/(0.200*20.0)=897 J(kg*K).