Q. How many grams of a liquid of specific heat $0.2$ at a temperature $40℃$ must be mixed with $100\ gm$ of a liquid of specific heat of $0.5$ at a temperature $20℃,$ so that the final temperature of the mixture becomes $32℃$
A) $175\ gm$
B) $300\ g$
C) $295\ gm$
D) $375\ g$
Explanation
(d) Temperature of mixture $\theta = \frac{m_{1}c_{1}\theta_{1} + m_{2}c_{2}\theta_{2}}{m_{1}c_{1} + m_{2}c_{2}}$
$\Rightarrow 32 = \frac{m_{1} \times 0.2 \times 40 + 100 \times 0.5 \times 20}{m_{1} \times 0.2 + 100 \times 0.5} \Rightarrow m_{1} = 375gm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.