Test Index

WAEC Physics 2012 Paper

Question : 21

Total: 47

^{-1}

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]

Solution:

Concept:

Heat lost by the hot copper equals heat gained by the melting ice at

0^\circ\mathrm{C}

Explanation:

The copper cools from

110^\circ\mathrm{C}

0^\circ\mathrm{C}

Heat lost by copper:

Q = m_{\mathrm{Cu}} \, c_{\mathrm{Cu}} \, \Delta\theta

Q = 20 \times 0.4 \times (110 - 0)

Q = 880 \, \mathrm{J}

This heat is used to melt ice without raising its temperature:

Q = m_{\mathrm{ice}} \, L

880 = m_{\mathrm{ice}} \times 330

m_{\mathrm{ice}} = \frac{880}{330}

m_{\mathrm{ice}} = 2.67 \, \mathrm{g}

Rounded to one decimal place:

m_{\mathrm{ice}} = 2.70 \, \mathrm{g}

Answer:

Option D:

2.70 \, \mathrm{g}

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