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Q.

A drilling machine is used to drill a hole in a 10 kg metal block. Power of the drilling machine is 1 KW and it is used for 5 minutes. If 40% of the work done by the drilling machine is converted into heat in the metal
block, the rise in its temperature is (specific heat - 0.12 cal gm^-1 ⁰C^–1 and J = 4.2 J/Cal)

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a

50°C

b

20⁰C

c

25°C

d

35°C

answer is B.

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Detailed Solution

$\begin{array}{l} m = 10 kg; p = 1 \times 10^{3} w; t = 5 \times 60 = 300 \sec \\ S = 0 .12 cal / gm -^{0} C; Δθ = ? \\ S = 120 cal / kg -^{\circ} C; J = 4 .2 J / cal \\ P = w / t (or) w = pt; w = 10^{3} \times 300 = 3 \times 10^{5} J \\ \frac{40}{100} w = JmSΔθ \\ Δθ = \frac{2}{5} \frac{w}{JmS} = \frac{2 \times 3 \times 10^{5}}{5 \times 4 .2 \times 10 \times 120} = \frac{100}{4 .2} = 23 .81 \\ Δθ = Nearly (25^{\circ} C) = 25^{\circ} C \end{array}$

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