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Thermal Properties of Matter

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Question : 12 of 22
Marks: +1, -0
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 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 =0.91 J g−1 K−1= 0.91\,\mathrm{J}\,\mathrm{g}^{-1}\,\mathrm{K}^{-1}.
Solution:  
Here, P=10 kW=104 W,P=10\,\mathrm{kW}=10^{4}\,\mathrm{W}, mass, m=8.0 kg=8×103 gm=8.0\,\mathrm{kg}=8 \times 10^{3}\,\mathrm{g}
rise in temperature, ΔT=?,\Delta T=?,
time, t=2.5 mint=2.5\,\mathrm{min}
=2.5×60=150 s=2.5 \times 60=150\,\mathrm{s}
Sp. heat, c=0.91 Jg−1 K−1c=0.91\,\mathrm{Jg}^{-1}\,\mathrm{K}^{-1}
Total energy, =P×t=104×150=P \times t=10^{4} \times 150
=15×105 J=15 \times 10^{5}\,\mathrm{J}
As, 50% of energy is lost,
∴\therefore Energy available, ΔQ=12×15×105\Delta Q = \frac{1}{2} \times 15 \times 10^{5}
=7.5×105 J=7.5 \times 10^{5}\,\mathrm{J}
As ΔQ=mcΔT\Delta Q = mc \Delta T
∴ΔT=ΔQmc\therefore \Delta T = \frac{\Delta Q}{m c}
=7.5×1058×103×0.91= \frac{7.5 \times 10^{5}}{8 \times 10^{3} \times 0.91}
=103∘C=103^{\circ}\mathrm{C}
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