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Q. 11.12 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^{o}/_{o} of power is used up in heating the machine itself or lost to the surrounding.  Spcific\; heat\; of\; aluminium=0.91\; j\; g^{-1}K^{-1}.

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Power of the drilling machine, P = 10 kW

Time. t = 2.5 min

Total energy dissipated E is

\\E=Pt\\ E=10\times 10^{3}\times 2.5\times 60\\ E=1.5\times 10^{6}J

Thermal energy absorbed by aluminium block is

\\\Delta Q=\frac{E}{2}\\ \Delta Q=\frac{1.5\times 10^{6}}{2}\\ \Delta Q=7.5\times 10^{5}J

Mass of the aluminium block, m = 8.0 kg

Specific heat of aluminium, c = 0.91 J g-1 K-1

Let rise in temperature be \Delta T

\\mc\Delta T=\Delta Q\\ \Delta T=\frac{\Delta Q}{mc}\\ \Delta T=\frac{7.5\times 10^{5}}{0.91\times 8\times 10^{3}}\\ \Delta T=103.02\ ^{o}C

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Sayak

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