JAMB Physics 2020 Paper

Concept:Specific heat capacity is the amount of heat energy required to raise the temperature of 1 kg of a substance by 1 K (or $1\,{}^{\circ}\text{C}$).Explanation:Heat is supplied uniformly at a power of $100\, \text{W}$ for $20\, \text{s}$.Total heat supplied is given by:$$Q = \text{power} \times \text{time} = 100 \times 20 = 2000\, \text{J}$$Mass of the liquid is $m = 1.0 \times 10^{-2}\, \text{kg}$.Temperature rise is $\Delta \theta = 5\,{}^{\circ}\mathrm{C}$.Using the heat formula:$$Q = mc\Delta\theta$$Substitute the known values:$$2000 = (1.0 \times 10^{-2}) \times c \times 5$$Simplify the right-hand side:$$2000 = 5 \times 10^{-2} c$$Solve for $c$:$$c = \frac{2000}{5 \times 10^{-2}} = \frac{2000}{0.05} = 40{,}000\, \text{J kg}^{-1}\text{K}^{-1}$$Thus the specific heat capacity is $4.0 \times 10^{4}\, \text{J kg}^{-1}\text{K}^{-1}$.Answer:$4.0 \times 10^{4}\, \text{J kg}^{-1}\text{K}^{-1}$ (Option C).