Question

A drilling machine of power P watts is used to drill a hole in a copper block of mass M kg. If the specific heat of the copper is $s\;Jkg^{-1}{}^{\circ}C^{-1}$ and 40% of the power is lost due to heating of the machine, the rise in the temperature of the block in T seconds will be (in ${}^{\circ}C$):
(A). $\dfrac{0.6PT}{Ms}$
(B). $\dfrac{0.6P}{MsT}$
(C). $\dfrac{0.4PT}{Ms}$
(D). $\dfrac{0.4P}{MsT}$

Answer 1

Hint: Recall that power is the energy consumed per unit time. In this case, obtain the expression for the energy consumed by the machine exclusive of the amount of power that is lost to machine heating. Then, obtain an expression for the amount of heat that the block can absorb to raise its temperature by $\Delta T$.
Now, the energy consumed by the machine is nothing but the amount of heat available for the copper block to absorb. Equate the two expressions you have so far and obtain the final relation.

Formula Used:
Power $P = \dfrac{E}{t}$
Heat energy $Q = mc\Delta T$

Complete step-by-step answer:
We are given that the power of the drilling machine is P.
We know that the power of the machine is nothing but the energy consumed by the machine in unit time. Therefore, power consumed by the machine in T seconds is given as:
$P = \dfrac{E}{T}$
The energy consumed by the machine in time T seconds will be:
$E = PT$
Now, we are given that 40% of the power is lost due to the heating of the machine. This means that 60% of the power is available for the drilling machine to consume. Therefore, the actual energy consumed by the machine will be:
$E = \dfrac{60}{100} \times PT = 0.6PT$
This fraction of energy is also available for the block to experience a rise in temperature $\Delta T$ as a result of heat transfer from the drilling machine.
$\Rightarrow Q = 0.6PT$
The amount of heat energy absorbed by the block of mass M kg, specific heat $s\;Jkg^{-1}{}^{\circ}C^{-1}$ to raise its temperature by $\Delta T$ is given as:
$Q = Ms\Delta T$
Now, equating the two equations above:
$0.6PT = Ms\Delta T \Rightarrow \Delta T = \dfrac{0.6PT}{Ms}$
Therefore, the correct option would be A. $\dfrac{0.6PT}{Ms}$

Note: Note that for our convenience we have assumed that all the power possessed by the machine after heating losses is imparted to the heating of the block. We are allowed to do so only here since nothing is specified contrarily. However, in case the question outrightly mentions the fraction of the machine power that is absorbed by the block, do not forget to account for that in your calculations.