Question

A ball of ice which is at $-20{}^{\circ }\text{C}$ temperature falls into a tub full of water which is at $52{}^{\circ }\text{C}$ temperature. Calculate the mass of ice, if mass of water in tub is $500\text{gm}$ and final temperature of mixture is $48{}^{\circ }\text{C}$.
Given : $S_{ice}=0.5{\text{cal/gm}}^{\circ }\text{C}$, $S_{water}=1{\text{cal/gm}}^{\circ }\text{C};L_{ice}=80\text{cal/gm}$

• A. $15\text{gms}$
• B. $16\text{gms}$
• C. $14.5\text{gms}$
• D. $10.5\text{gms}$

Answer 1

The correct option is C $14.5\text{gms}$

Total heat released by water is,
$Q_{lost}=ms\Delta T$
$=500\times 1\times 4$
$=2000\text{Cal}$

Heat gained by ice is, $Q_{gain}$,
Assuming mass of ice to be $m$

$Q_{gain}=$ Heat required to change ice from $-{20}^{\circ }C$ to $0^{\circ }C$
$+$
Heat require to change state from ice to water
$+$
Heat required to change temperature from $0^{\circ }C$
Water to ${48}^{\circ }C$ water.

$Q_{gain}=m{S}_{ice}\left(20\right)+m{L}_{ice}+m{S}_{water}\left(48\right)$
$=m[\frac{1}{2}\times 20+80+48]$
$=138m\text{cal}$

By law of conservation of energy
$Q_{lost}=Q_{gain}$
$\Rightarrow 2000=138m$
$\Rightarrow m\approx 14.5\text{gms}$

Hence, $\left(\text{C}\right)$ is the correct answer.

Why this question ? This question gives you concept related to specific heat, latent heat during change of temperature and change of phase.