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JAMB Physics 2023 Paper

Question : 21

Total: 65

-10^oC

10^oC

10^5

Jkg^-1

Jkg^-1

k^-1

Solution:

Concept:

Heating the ice from

-10^\circ\text{C}

10^\circ\text{C}

happens in three stages: warming ice, melting it, then warming the resulting water.

Explanation:

Stage 1: Raise ice from

-10^\circ\text{C}

0^\circ\text{C}

using

c_{\text{ice}} = 2100\,\text{Jkg}^{-1}\text{K}^{-1}

H_1 = mc_{\text{ice}}\Delta T = 0.02 \times 2100 \times (0 - (-10)) = 420\,\text{J}

Stage 2: Melt the ice at

0^\circ\text{C}

using the specific latent heat of fusion.

H_2 = mL_f = 0.02 \times 3.34 \times 10^5 = 6680\,\text{J}

Stage 3: Warm the melted ice (water) from

0^\circ\text{C}

10^\circ\text{C}

using

c_{\text{water}} = 4200\,\text{Jkg}^{-1}\text{K}^{-1}

H_3 = mc_{\text{water}}\Delta T = 0.02 \times 4200 \times 10 = 840\,\text{J}

Total heat required:

H = H_1 + H_2 + H_3 = 420 + 6680 + 840 = 7940\,\text{J}

Answer:

7940\,\text{J}

, so the correct option is D.

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