How many calories are absorbed by a pot of water with a mass of 500 g in order to raise the temperature from 20 c to 30 c

Q = 5 kCal

Solution:

The equation used for this problem is as follow,

Q = m Cp ΔT

Where;

Q = Heat = ?

m = mass = 500 g

Cp = Specific Heat Capacity = 4.18 J. g⁻¹.°C⁻¹

ΔT = Change in Temperature = 30 °C - 20 °C = 10 °C

Putting values in eq. 1,

Q = 500 g * 4.18 J. g⁻¹.°C⁻¹ * 10 °C

Q = 20900 J

Q = 4.99 kCal ≈ 5 kCal