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JAMB Physics 2011 Paper

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Question : 20

Total: 44

A block of aluminum is heated electrically by a 30 w heater if the temperature rises by 10°c in 5 minutes, the heat capacity of the aluminum is

900 JK^-1
90 JK^-1
100 JK^-1
200 JK^-1

Solution:

Concept:

Heat capacity is the amount of heat needed to raise the temperature of the object by

1\,\mathrm{K}

.

Explanation:

The heater supplies energy at

30\,\mathrm{W}

for

5

minutes.

Convert time to seconds:

t = 5 \times 60 = 300\,\mathrm{s}

.

Heat supplied:

Q = Pt = 30 \times 300 = 9000\,\mathrm{J}

.

Temperature rise:

\Delta T = 10^\circ\mathrm{C} = 10\,\mathrm{K}

.

Using

Q = C\Delta T

, where

C

is the heat capacity:

C = \frac{Q}{\Delta T} = \frac{9000}{10} = 900\,\mathrm{J\,K^{-1}}

Answer:

900\,\mathrm{J\,K^{-1}}

Correct option: A.

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