Specific heat chemistry problem

Question

A 34.86 g sample of a substance is initially at 27.5 °C. After absorbing 1699 J of heat, the temperature of the substance is 148.1 °C. What is the specific heat (c) of the substance?

Answer 1

c = (1/m) ΔQ / ΔT

 

m = 34.86 g = 0.03486 kg

 

ΔQ  = 1699 J

 

ΔT = 148.1°C - 27.5°C = 120.6°C

 

Note the temperature difference is the same in ºC and ºK.

 

c = (1/0.03486 kg) (1699 J) / (120.6°C)

 

c = 404.1 J / (kg °C)