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 10\(^5\) Jkg\(^{-1}\) specific heat capacity of water 4.2 x 10\(^3\) J kg\(^{-1}\) K\(^{-1}\)]
A. 4.80 kJ B. 20.16 kJ C. 40.32 kJ D. 22.17 kJ
Correct Answer: C
Explanation
Heat needed to change 60g (0.06kg) of ice at 0\(^o\)C to water at 0\(^o\) = 0.06 x 3.36 x 10\(^5\)J = 0.2016 x 10\(^5\) J Heat required to raise the temp. of 60g (0.06kg) of water from 0\(^o\)C to 80\(^o\)C = 0.06 x 4.2 x 10\(^3\) x (80.0)J = 0.06 x 4.2 x 10\(^3\) x 80 = 20.16 x 10\(^3\) = 0.2016 x 10\(^5\) Total heat needed = (0.2016 + 0.2016) 10\(^5\)J = 0.4032 x 10\(^5\) J = 40.32KJ
