5 g of water at 30° C and 5 g of ice at -20° 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 °, and latent heat of ice is 80 cal / g.

Question

5 g of water at 30° C and 5 g of ice at -20° 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 °, and latent heat of ice is 80 cal / g. (A) 0° C (B) 10° C (C) -30° C (D) >10° C

Tardigrade Admin · Accepted Answer

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 Δ 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 = - 3 0^{\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

(0^{\circ} C)

, the temperature of system will not fall below

0^{\circ} C

.

$0^{\circ} C$

$1 0^{\circ} C$

$- 3 0^{\circ} C$

$> 1 0^{\circ} C$

Q. 5g of water at 30∘C and 5g of ice at −20∘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 1cal/gC∘, and latent heat of ice is 80cal/g.

0∘C

10∘C

−30∘C

>10∘C

$0^{\circ} C$

$1 0^{\circ} C$

$- 3 0^{\circ} C$

$> 1 0^{\circ} C$