JAMB Physics 2021 Paper

Concept:Vaporizing water that is below its boiling point requires two energy steps: first raising its temperature to $100^\circ\text{C}$, then supplying the latent heat to change it to vapour.Explanation:Given $m = 50\,\text{g}$, $c = 4.23\,\text{J g}^{-1}\text{K}^{-1}$, $L = 2260\,\text{J g}^{-1}$, and initial temperature $=80^\circ\text{C}$.Temperature rise needed: $\Delta T = 100 - 80 = 20^\circ\text{C}$.Total heat energy is $Q = mc\Delta T + mL$.Sensible heat to reach boiling point: $Q_1 = 50 \times 4.23 \times 20 = 4230\,\text{J}$.Latent heat of vaporization: $Q_2 = 50 \times 2260 = 113000\,\text{J}$.Total heat energy: $Q = 4230 + 113000 = 117230\,\text{J}$.This result is not exactly any of the listed choices.The value $130000\,\text{J}$ would require $\Delta T = 80^\circ\text{C}$, which corresponds to starting from $20^\circ\text{C}$, not $80^\circ\text{C}$.Answer:For the question as stated, the heat required is $117230\,\text{J}$, so none of the options A–D is correct.