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

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