Question

A drilling machine of power $P$ watt is used to drill a hole in copper block of mass $M$ kg. If the specific heat of copper is $sJk{g}^{-1}{}^o{C}^{-1}$ 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 ${}^{o}C$)

• A. $\frac{0.6PT}{Ms}$
• B. $\frac{0.6P}{MsT}$
• C. $\frac{0.4PT}{Ms}$
• D. $\frac{0.4P}{MsT}$

Answer 1

Answer: (A)

$\frac{0.6PT}{Ms}$

Power of machine = $P$;
Energy used by machine in time '$T$' = $PT$
Out of which $40\%$ is wasted
i.e. Energy used on the block = $0.6PT=Q$
This energy is used to heat the block
The rise in temperature is given by:
$Q=mS\Delta T$ , $M$ is the mass of the block and $s$ is the specific heat capacity.
i.e.
$\Delta T=\frac{Q}{Ms}=\frac{0.6PT}{Ms}$
Hence the answer is option A