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JAMB Physics 2020 Paper

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Question : 30

Total: 38

Heat is supplied uniformly at the rate of 100W to 1.0 x 10

^{-2}

kg of a liquid for 20 seconds. If the temperature of the liquid rises by 5

^{\circ}

C, then the specific heat capacity of the liquid is

2.0 x 10 $^{2}$ Jkg $^{-1}$ K $^{-1}$
2.0 x 10 $^{2}$ Jkg $^{-1}$
4.0 x 10 $^{4}$ Jkg $^{-1}$ K $^{-1}$
4.0 x 10 $^{4}$ Jkg $^{-1}$

Solution:

Heat energy = Electrical Energy

MC $\theta$ = power x time

5 x 1 x 10 $^{-2}$ x C = 100 x 20

C = $\frac{200}{5 \times 10^{2}}$

= 4 x 10 $^{4}$ J/kg/ $^{\circ}\mathrm{C}$

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