JAMB Physics 2021 Paper

Concept:Total heat extracted equals heat removed during cooling plus heat removed during freezing.Explanation:Given: mass $m = 0.5\ \mathrm{kg}$, temperature change $\Delta\theta = 10^\circ\mathrm{C} - 0^\circ\mathrm{C} = 10^\circ\mathrm{C}$, heat extracted $H = 88000\ \mathrm{J}$, specific heat capacity $c = 4200\\ \mathrm{J\,kg^{-1}\,{}^\circ C^{-1}}$.Heat removed while cooling water from $10^\circ\mathrm{C}$ to $0^\circ\mathrm{C}$: $mc\Delta\theta = 0.5 \times 4200 \times 10 = 21000\ \mathrm{J}$.Heat removed during freezing at $0^\circ\mathrm{C}$: $mL = 0.5L$.Total heat extracted: $H = mc\Delta\theta + mL$.Substitute and rearrange: $88000 = 21000 + 0.5L$.Thus $0.5L = 88000 - 21000 = 67000$.So $L = \frac{67000}{0.5} = 134000\ \mathrm{J\,kg^{-1}} = 134\ \mathrm{kJ\,kg^{-1}}$.Answer:The specific latent heat of fusion of ice is $134.0\ \mathrm{kJ\,kg^{-1}}$, which is option C.