Heat And Thermodynamics Ques 28
- $300 g$ of water at $25^{\circ} C$ is added to $100 g$ of ice at $0^{\circ} C$. The final temperature of the mixture is..${ }^{\circ} C$.
$(1989,2 M)$
Show Answer
Solution:
Formula:
- Heat liberated when $300 g$ water at $25^{\circ} C$ goes to water at $0^{\circ} C: Q=m s \Delta \theta=(300)(1)(25)=7500 cal$
From $Q=m L$, this much heat can melt mass of ice given by
$$ m=\frac{Q}{L}=\frac{7500}{80}=93.75 g $$
i.e. whole ice will not melt.
Hence, the mixture will be at $0^{\circ} C$.
Mass of water in mixture
$$ \begin{aligned} & =300+93.75 \\ & =393.75 g \text { and } \end{aligned} $$
Mass of ice in mixture
$$ =100-93.75=6.25 g $$
BETA
