Question

An amount of water of mass $20g$, at $0^{0}C$ is mixed with $40g$ of water at ${100}^{0}C$. Final temperature of mixture is:

• A. $-{20}^{0}C$
• B. ${66.67}^{0}C$
• C. $5^{0}C$
• D. $0^{0}C$

Answer 1

The correct option is B ${66.67}^{0}C$

Heat lost by hot water $=$ Heat gained by cold water
$m_1.s.\Delta T$ = $m_2.s.\Delta T$
$40\left(1\right)(100-T)=20\left(1\right)(T-0)$
$200-2T=T$
$3T=200$
$T={66.67}^{o}C$