Question

The specific heat of a certain type of cooking oil is 0.418 cal/g °C). How much heat energy is needed to raise the temperature of 1.152 kg of this oil from 23.0 °C to 60.0 °C?

Answer 1

Q = m×c×ΔT

• Q is the change in thermal energy in J
• m is the mass in kg
• c is the specific heat capacity in J/kg°C
• ΔT is the temperature change in °C

cal c = 0.418 - = 0.418 x 4.186 ) - = 1747.24 J/K 3°C (10-3K9)°C go

ΔΤ = 60 – 23 = 37°C

m = 1.152 Kg

Q = 74474.35 J = 74.47 kJ