Question

A drilling machine of power P watts is used to drill a hole in copper block of mass M kg. If the specific heat of copper is s J/kg-C 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 °C) ?

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

Answer 1

Answer: (D)

$\frac { 0.6 \mathrm { PT } } { \mathrm { Ms } }$

Answer 2

useful power = 0.6 P

\ energy consumed in T sec. = 0.6 PT

Now, MsDT = 0.6 PT

\ DT = $\frac { 0.6 \mathrm { PT } } { \mathrm { Ms } }$