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A 10 kW drilling machine is used for 100 sec to drill a bore in a small aluminum block of mass 8 kg. If 25% of energy is lost and the remaining part of energy is used for heating the block, then the rise of temperature of the block is (specific heat of aluminum = 900 J/kg/K)

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52°C

47°C

377°C

104°C

answer is D.

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Detailed Solution

$\begin{array}{l} 8 \times 900 \times Δθ = \frac{75}{100} \times (10 \times 10^{3} \times 100) \\ \Rightarrow Δθ = 104 {.16}^{\circ} C . \end{array}$

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