Question

When m grams of water at ${10}^{\circ }C$ is mixed with m grams of ice at $0^{\circ }C$, which of the following statements are false?

• A. The temperature of the system will be given by the equation $m\times 80+m\times 1\times (T-0)=m\times 1\times (10-T)$
• B. Whole of ice will melt and temperature will be more that $0^{\circ }C$ but lesser than ${10}^{\circ }C$
• C. Whole of ice will melt and temperature will be $0^{\circ }C$
• D. Whole of ice will not melt and temperature will be $0^{\circ }C$

Answer 1

Answer: (A), (B), (C)

Whole of ice will melt and temperature will be $0^{\circ }C$

$Q_1$ = heat given out by water if it was to cool up to $0^{\circ }C$
$=m\times 1{\text{cal/g}}^{\circ }C\times (10-0)=10\text{m cal}$
$Q_2$ = heat required by ice to melt completely into water at $0^{\circ }C$
$=m\times L_F=m\times 80\text{cal/g}=80\text{m cal}$
As $Q_1<Q_2$, equilibrium temperature will be $0^{\circ }C$ and whole of ice will not melt.

Why this question? Tip: Check whether the maximum possible heat supplied by the hot body is enough to change the phase of cooler body first, if not - it will save us the calculation for finding final temperature.