Ice at 0^oC is added to 200,g of water initia | vedclass

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(A) Let $W$ be the water equivalent of the flask in grams and $L$ be the latent heat of fusion of ice in $cal/g$.Step $1$: Heat lost by water and flas

Vedclass · Accepted Answer

$3.8 	imes 10^5\,J/kg$

Vedclass · Answer

(A) Let $W$ be the water equivalent of the flask in grams and $L$ be the latent heat of fusion of ice in $cal/g$.Step $1$: Heat lost by water and flask = Heat gained by ice.$(200 + W)(70 - 40) = 50L + 50(40 - 0)$$30(200 + W) = 50L + 2000$$6000 + 30W = 50L + 2000 \implies 50L - 30W = 4000 \implies 5L - 3W = 400$ ---$(1)$Step $2$: For the second addition,total mass of water is $(200 + 50 + 80) = 330\,g$.Heat lost by water and flask = Heat gained by ice.$(200 + 50 + W)(40 - 10) = 80L + 80(10 - 0)$$30(250 + W) = 80L + 800$$7500 + 30W = 80L + 800 \implies 80L - 30W = 6700 \implies 8L - 3W = 670$ ---$(2)$Step $3$: Subtract $(1)$ from $(2)$:$(8L - 3W) - (5L - 3W) = 670 - 400$$3L = 270 \implies L = 90\,cal/g$.Step $4$: Convert to $J/kg$:$L = 90 	imes 4.184 	imes 10^3\,J/kg \approx 3.76 	imes 10^5\,J/kg \approx 3.8 	imes 10^5\,J/kg$.

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