Question

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 61.0°C, what is the specific heat of the substance?
Use q equals m C subscript p Delta T..
0.171 J/(gi°C)
0.548 J/(gi°C)
15.9 J/(gi°C)
86.6 J/(gi°C)

Answer 1

Use q = m c ΔT, so c = q/(m ΔT).

ΔT = 61.0 − 32.0 = 29.0 °C.
m = 0.158 kg = 158 g.

c = 2510 J / (158 g × 29.0 °C) = 2510 / 4582 ≈ 0.548 J/(g·°C).

Answer: 0.548 J/(g·°C).