Calculate the quantity of heat needed to change the temperature of 60g of ice at 0 °C to 80 °C. [specific latent heat of fusion of ice= 3.36 x 105 Jkg‾¹ specific heat capacity of water 4.2 x 10³ J kg‾¹ K‾¹]

A. 4.80 kJ
B. 20.16 kJ
C. 40.32 kJ
D. 22.17 kJ

Correct Answer:

Option C = 40.32 kJ

Explanation

Heat needed to change 60g (0.06kg) of ice at 0^oC to water at 0o

= 0.06 x 3.36 x 10⁵J

= 0.2016 x 10⁵ J

Heat required to raise the temp. of 60g (0.06kg) of water from 0^oC to 80^oC

= 0.06 x 4.2 x 10³ x (80.0) J

= 0.06 x 4.2 x 10³ x 80

= 20.16 x 10³

= 0.2016 x 10⁵

Total heat needed = (0.2016 + 0.2016) x 10⁵J

= 0.4032 x 10⁵ J

= 40.32KJ