$$10$$ 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)

Question

Infinity Learn · Accepted Answer

Let T be the temperature when thermal equilibrium has reached.Heat gained by ice to be converted to water at $$00C=mL+m$$×c×(T$$−$$0)=10$$×$$80+10$$×$$1$$×THeat lost by tumbler and its contents $$=$$ $$55$$ x $$(40$$ $$-$$ $$T)Using$$ principle of calorimetry, heat gained $$=$$ heat $$lost10$$×$$80+10$$×$$1$$×$$T=55$$×$$(40$$−$$T)⇒ $$65T=2200$$−$$800=1400$$⇒ $$T=140065$$≈$$21.5$$∘C≈$$22$$∘C

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)

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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)

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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)