A drilling machine of 10 , kW power is used t | vedclass

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Question

(A) Power of the drilling machine $P = 10 \, kW = 10^4 \, W$.Time $t = 2.5 \, minutes = 2.5 	imes 60 = 150 \, s$.Total energy supplied $E = P 	imes

Vedclass · Accepted Answer

$103$

Vedclass · Answer

(A) Power of the drilling machine $P = 10 \, kW = 10^4 \, W$.Time $t = 2.5 \, minutes = 2.5 	imes 60 = 150 \, s$.Total energy supplied $E = P 	imes t = 10^4 	imes 150 = 1.5 	imes 10^6 \, J$.Since $50 \%$ of the power is lost,the energy absorbed by the aluminium block is $Q = 50 \% 	ext{ of } E = 0.5 	imes 1.5 	imes 10^6 = 7.5 	imes 10^5 \, J$.The mass of the block $m = 8 \, kg = 8000 \, g$.The specific heat capacity of aluminium $s = 0.91 \, J/g \cdot ^\circ C$.Using the formula $Q = ms \Delta T$,we have:$7.5 	imes 10^5 = 8000 	imes 0.91 	imes \Delta T$.$7.5 	imes 10^5 = 7280 	imes \Delta T$.$\Delta T = \frac{750000}{7280} \approx 103.02 \, ^\circ C$.Therefore,the rise in temperature is approximately $103 \, ^\circ C$.

List-$I$	List-$II$
$(A)$ Heat capacity of body	$(I)$ $J kg^{-1}$
$(B)$ Specific heat capacity of body	$(II)$ $J K^{-1}$
$(C)$ Latent heat	$(III)$ $J kg^{-1} K^{-1}$
$(D)$ Thermal conductivity	$(IV)$ $J m^{-1} K^{-1} s^{-1}$

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