Solution:
Heat lost by tumbler = Heat gained by ice
$55 cdot (40 - T) = 10 cdot 80 + 10 cdot T$
$2200 - 55T = 800 + 10T Rightarrow 65T = 1400 Rightarrow T approx 21.5^circ text{C} approx 22^circ text{C}$
$10 text{ gm}$ of ice cubes at $0^circ text{C}$ are released in a tumbler (water equivalent $55 text{ g}$) at $40^circ text{C}$. Assuming that negligible heat is taken from the surroundings the temperature of water in the tumbler becomes nearly ($L = 80 text{ cal/g}$):
(1988)
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