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JAMB Physics 2018 Paper

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Question : 53

Total: 87

An electric heating coil rated at 1Kw is used to heat 2kg of water for 2 minutes. The initial water temperature is 30

^o

C. Taking the specific heat of the water as 4,000Jkg

^{-1}

and neglecting that of the container, the final water temperature is

15.0 $^o$ C
45.0 $^o$ C
52.5 $^o$ C
60.0 $^o$ C

Solution:

Concept:

Electrical energy supplied by the coil is completely used to heat the water, so

P t = m c \Delta \theta

.

Explanation:

Power of the coil:

P = 1\,\text{kW} = 1000\,\text{W}

.

Time of heating:

t = 2\,\text{min} = 120\,\text{s}

.

Energy supplied:

E = P t = 1000 \times 120 = 120{,}000\,\text{J}

.

Heat absorbed by water:

Q = m c (\theta_2 - \theta_1)

.

Substitute

m = 2\,\text{kg}

,

c = 4000\,\mathrm{J\,kg^{-1}\,{}^\circ C^{-1}}

, and

\theta_1 = 30^\circ\text{C}

:

120{,}000 = 2 \times 4000 \times (\theta_2 - 30)

120{,}000 = 8000\,(\theta_2 - 30)

\theta_2 - 30 = \frac{120{,}000}{8000} = 15

\theta_2 = 30 + 15 = 45^\circ\text{C}

Answer:

B.

45.0^\circ\text{C}

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