Question

When $50\, g$ of water at $10^{\circ} C$ is mixed with $50\, g$ of water at $100^{\circ} C$, the resultant temperature is

• A. $80^{\circ} C$
• B. $55^{\circ} C$
• C. $25^{\circ} C$
• D. $45^{\circ} C$

Answer 1

Answer: (B)

If resulting temperature of the mixture is $T^{\circ} C$ then Heat gained by $50\, g$ water at $10^{\circ} C$,
$H_{\text {gain }}=m c \Delta T=50 c(T-10)$
Heat lost by $50\, g$ of water at $100^{\circ} C$
$H_{\text {loss }}=m c \Delta T=50 c(100-T)$
According to principle of calorimetry,
Heat gain $=$ Heat loss
$\Rightarrow H_{\text {gain }}=H_{\text {loss }}$
$\Rightarrow 50 c(T-10)=50 c(100-T)$
$\Rightarrow 2 T=110$
$T=55^{\circ} C$