Q.

300 grams of water at 25∘C is added to 100 grams of ice at 0∘C , the final temperature of the mixture is

Moderate

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By Expert Faculty of Sri Chaitanya

−53 ∘C

−52 ∘C

-5oC

0∘C

answer is 4.

(Detailed Solution Below)

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Detailed Solution

Wateramount of heat lost by water is dQ1=mSdθ here m is mass of water ; S is specific heat of water; dθ is change in temperature of water when water looses its temperature from 250C to 00C, amount of heat lost is dQ1 dQ1=300×1×25=7500 calorieFor Iceamount of heat required to convert from ice to water is dQ2=mL where m is mass of ice ; L = latent heat of fusion of ice=80 cal g-1 dQ2=100×80=8000 caldQ2>dQ1i.e., Heat required is greater than heat availableStill ice remains= 0oC

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