Q. An amount of water of mass 20 g at 0$℃\ $ is mixed with 40 g of water 10$℃,$ final temperature of the mixture is
A) 5$℃$
B) 0$℃$
C) 20$℃$
D) 6.66$℃$
Explanation
(d) Let $\theta\ be\ the$ temperature of the mixture.
Heat gained by water at $0℃$ = Heat lost by water at 10$℃$
$c\ m_{1}\ (\theta - 0) = \ c\ m_{2}(10 - \theta)$
$\theta = \frac{400}{60} = 6.66℃$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.