First slide

Questions based on phase change

Question

10 g of ice cubes at 0⁰C are released in a tumbler (water equivalent 55 g) at 40⁰C. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (Take, L = 80 cal/g)

Moderate

A

31⁰C
B

22⁰C
C

19⁰C
D

15⁰C

Solution

Let T be the temperature when thermal equilibrium has reached.
Heat gained by ice to be converted to water at 0⁰C
$= m L + m \times c \times (T - 0) = 10 \times 80 + 10 \times 1 \times T$
Heat lost by tumbler and its contents = 55 x (40 $-$ T)
Using principle of calorimetry, heat gained = heat lost
$10 \times 80 + 10 \times 1 \times T = 55 \times (40 - T)$
$\Rightarrow 65 T = 2200 - 800 = 1400$
$\Rightarrow T = \frac{1400}{65} \approx {21.5}^{\circ} C \approx 22^{\circ} C$

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