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A $10 \mathrm{~kW}$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0 \mathrm{~kg}$. How much is the rise in temperature of the block in $2.5$ minutes, assuming $50 \%$ of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium $=\mathbf{0 . 9 1 J} / \mathrm{g}^{-1} \mathrm{~K}^{-1}$
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Total energy $=P \times t=10^4 \mathrm{~W} \times 150 \mathrm{~s}=15 \times 10^5 \mathrm{~J}$ As $50 \%$ of the heat is lost,
$\therefore \quad$ Energy available $=\frac{50}{100} \times 15 \times 10^5$
$=7.5 \times 10^5 \mathrm{~J}$
$\because Q=m C \Delta T=8 \times 10^3 \times 0.91 \times \Delta T$
$\therefore \quad 7.5 \times 10^5=8 \times 10^3 \times 0.91 \times \Delta T$
$\Rightarrow \Delta T=\frac{7.5 \times 10^5}{8 \times 10^3 \times 0.91}=103^{\circ} \mathrm{C}$
$\therefore \quad$ Energy available $=\frac{50}{100} \times 15 \times 10^5$
$=7.5 \times 10^5 \mathrm{~J}$
$\because Q=m C \Delta T=8 \times 10^3 \times 0.91 \times \Delta T$
$\therefore \quad 7.5 \times 10^5=8 \times 10^3 \times 0.91 \times \Delta T$
$\Rightarrow \Delta T=\frac{7.5 \times 10^5}{8 \times 10^3 \times 0.91}=103^{\circ} \mathrm{C}$
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