Question

A 34.34 g sample of a substance is initially at 26.7 °C. After absorbing 2205 J of heat, the temperature of the substance is 152.1 °C. What is the specific heat (c) of the substance?

Answer 1

The specific heat of the substance is c= 512.04 J/kg K

Step-by-step explanation:

ΔQ= 2205 J

m= 0.03434 kg

ΔT= 125.4 ºC

ΔQ= m * c * ΔT

c= ΔQ / (m * ΔT)

c= 512.04 J/Kg K