Q. A calorimeter of mass $0.2\ kg$ and specific heat $900\ J/kg$-$K$. Containing $0.5\ kg$ of a liquid of specific heat $2400\ J/kg$-$K$. Its temperature falls from $60℃$ to $55℃$ in one minute. The rate of cooling is
A) $5\ J/s$
B) $15\ J/s$
C) $100\ J/s$
D) $115\ J/s$
Explanation
(d) Rate of cooling (here it is rate of loss of heat)
$\frac{dQ}{dt} = (mc + W)\frac{d\theta}{dt} = \left( m_{l}c_{l} + m_{c}c_{c} \right)\frac{d\theta}{dt}$
$\Rightarrow \frac{dQ}{dt} = (0.5 \times 2400 + 0.2 \times 900)\left( \frac{60 - 55}{60} \right) = 115\ \frac{J}{s}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.