First slide

Questions on calorimetry without phase change

Question

10 gm of ice at -20^oC is dropped into a calorimeter containing 10 gm of water at 10^oC; the specific heat of water is twice that of ice. When equilibrium is reached, the calorimeter will contain

Moderate

A

20 gm of water
B

20 gm of ice
C

10 gm ice and 10 gm water
D

5 gm ice and 15gm water

Solution

$\begin{array}{l} Q_{1} = 10 \times 1 \times 10 = 100 cal \\ Q_{2} = 10 \times 0 .5 [0 - (- 20)] + 10 \times 80 \\ = (100 + 800) cal = 900 cal \end{array}$
As Q₁ < Q₂, so ice will not completely melt and final temperature = 0^oC.
As heat given by water in cooling upto 0^oC is only just sufficient to increase the temperature of ice from -20^oC to 0^oC hence mixture in equilibrium will consist of 10 gm of ice and 10 gm of water, both at $0^{o}$ C.

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