Question

$100\, g$ of ice (latent heat $80 \,cal\, g ^{-1}$, at $0^{\circ} C$ ) is mixed with $100 \,g$ of water (specific heat $1 \,cal \,g ^{-1}{ }^{\circ} \,C ^{-1}$ ) at $80^{\circ} \,C$. The final temperature of the mixture will be

• A. $0^{\circ} C$
• B. $40^{\circ} C$
• C. $80^{\circ} C$
• D. $<0^{\circ} C$

Answer 1

Answer: (A)

$100 \,g$ ice $0^{\circ} C +100\, g$ water $80^{\circ} C$
$Q_{\text {required }}=m L $
if $\Delta T=80 $
$\left[Q_{f}\right.$ becomes $\left.=0^{\circ}\right]$
$=100 \times 80 $
$Q_{\text {available }}=m C \Delta T$
$=8000\, cal $
$=100 \times 1 \times 80=8000 \,cal$
[Required to melt whole ice]
Since $Q_{\text {required }}=Q_{\text {avallable }}$
$\therefore $ Whole system will be water at equillibrium and temperature would be $0^{\circ} C$