• # question_answer A 10k W drilling machine is used to drill a bore in an aluminium block of mass 8.0 kg. Block is worked on by machine for 2.5 min to drill a hole and 50% of power is used up in heating the aluminium block. Specific heat of aluminium is $0.91J{{g}^{-1}}{{K}^{-1}}$. Rise in temperature of block due to drilling will be A) $100{}^\circ C$ B) $103{}^\circ K$ C) $103{}^\circ C$ D) $50{}^\circ C$

 Here, power $=110kW={{10}^{4}}W$, mass $m=8.0kg=8\times {{10}^{3}}g$,time$t=2.5\min =2.5\times 60=150s$and specific heat $c=0.91J{{g}^{-1}}{}^\circ C$ Total energy =power $\times$Time =${{10}^{4}}\times 150\operatorname{J}=$ As, $50%$of energy is lost. Hence, thermal energy available, $\Delta Q=\frac{1}{2}\times 15\times {{10}^{5}}=7.5\times {{10}^{5}}J$ As, $\Delta Q=\operatorname{mc}\Delta \operatorname{T}$$\Rightarrow$$\Delta T=\frac{\Delta Q}{mc}=\frac{7.5\times {{10}^{5}}}{8\times {{10}^{3}}\times 0.91}=103{}^\circ C$