Question

$5 \,g$ of water at $30^{\circ} C$ and $5 \,g$ of ice at $-20^{\circ} C$ are mixed together in a calorimeter. Find the final temperature of the mixture. Assume water equivalent of calorimeter to be negligible, sp. heats of ice and water are $0.5$ and $1 \,cal / g C ^{\circ}$, and latent heat of ice is $80\, cal / g$.

• A. $0^{\circ} C$
• B. $10^{\circ} C$
• C. $-30^{\circ} C$
• D. $>10^{\circ} C$

Answer 1

Answer: (A)

Here ice will absorb heat while hot water will release it. So if $T$ is the final temperature of the mixture, heat given by water
$Q_{1}=m c \Delta T=5 \times 1 \times(30-T)$
And heat absorbed by ice
$Q_{2}=5 \times(1 / 2)[0-(-20)]+5 \times 80+5 \times 1(T-0)$
So, by principle of calorimetry $Q_{1}=Q_{2}$, i.e.,
$150-5 T=450+5 T $
$T=-30^{\circ} C$
Which is impossible as a body cannot be cooled to a temperature below the temperature of cooling body. The physical reason for this discrepancy is the heat remaining after changing the temperature of ice from $-20$ to $0^{\circ} C$ with some ice left unmelted and we are taking it for granted that heat is transferred from water at $0^{\circ} C$ to ice at $0^{\circ} C$ so that temperature of system drops below $0^{\circ} C$.
However, as heat cannot flow from one body (water) to the other (ice) at same temperature $\left(0^{\circ} C \right)$, the temperature of system will not fall below $0^{\circ} C$.