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Physics Question on thermal properties of matter

A 10 kW10\, kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg8 \,kg. Find the rise in temperature of the block in 2.52.5 minutes, assuming 50%50\% power is used up in heating the machine itself or lost to the surroundings. (Specific heat of aluminium =0.91 J g−1 ∘C−1= 0.91\, J \,g^{-1} \,^{\circ}C^{-1})

A

100 ∘C100 \,^{\circ}C

B

103 ∘C103 \,^{\circ}C

C

150 ∘C150 \,^{\circ}C

D

155 ∘C155 \,^{\circ}C

Answer

103 ∘C103 \,^{\circ}C

Explanation

Solution

Here, P=10 kW=104 WP = 10\, kW = 10^4\, W, m=8 kgm = 8 \,kg time, t=2.5 minute=2.5×60=150 st = 2.5\, minute = 2.5 \times 60 = 150 \,s Specific heat, s=0.91 J g−1 C−1s = 0.91\, J \,g^{-1}\, C^{-1} Total energy =P×t=104×150=15×105 J= P \times t = 10^4 \times 150 = 15 \times 10^5 \,J As 50%50\% of energy is lost, ∴\therefore Energy available, ΔQ=12×15×105=7.5×105 J\Delta Q=\frac{1}{2}\times15\times10^{5}=7.5 \times10^{5}\,J As ΔQ=msΔT\Delta Q=ms\Delta T ΔT=ΔQms=7.5×1058×103×0.91≈103 ∘C\Delta T=\frac{\Delta Q}{ms}=\frac{7.5 \times 10^{5}}{8 \times 10^{3} \times 0.91}\approx103 \,^{\circ}C