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WAEC Physics 2010 Paper

Question : 21

Total: 50

An electric drill rated 400W is used to drill a hole in copper of mass 400g in 20s. Calculate the rise in temperature if all the heat produced is absorbed by the copper. [specific heat capacity of copper = 400Jkg^-1K^-1]

100^oC
75^oC
50^oC
45^oC

Validate

Solution:

Concept:

The electrical energy supplied to the drill is converted into heat, and this heat is completely absorbed by the copper block, raising its temperature.

Explanation:

Electrical energy converted to heat is given by

Q = Pt

Here,

P = 400\,W

and

t = 20\,s

, so

Q = 400 \times 20 = 8000\,J

Mass of copper is

m = 400\,g = 0.4\,kg

Specific heat capacity of copper is

c = 400\,Jkg^{-1}K^{-1}

The heat absorbed is

Q = mc\Delta\theta

Therefore,

\Delta\theta = \frac{Q}{mc} = \frac{8000}{0.4 \times 400} = \frac{8000}{160} = 50^{\circ}C

Answer:

The rise in temperature is

50^{\circ}C

, which is option C.

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