Question

What is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°C to 70.0°C when 2,520 J of heat is applied?
Use q equals m C subscript p Delta T..
0.00420 J/(gi°C)
0.00661 J/(gi°C)
238 J/(gi°C)
252 J/(gi°C)

Answer 1

ΔT = 70.0 − 10.0 = 60.0 °C.
c = q / (m ΔT) = 2520 J / (10.0 kg × 60.0 °C) = 4.20 J/(kg·°C) = 0.00420 J/(g·°C).

Answer: 0.00420 J/(g·°C).