When 100 g of boiling water at 100^{circ} C | vedclass

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(D) Let $W$ be the water equivalent of the calorimeter in $g$. Step $1$: Heat lost by hot water = Heat gained by cold water + Heat gained by calorimet

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$0.1$

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(D) Let $W$ be the water equivalent of the calorimeter in $g$. Step $1$: Heat lost by hot water = Heat gained by cold water + Heat gained by calorimeter. $100 	imes 1 	imes (100 - 20) = 300 	imes 1 	imes (20 - 10) + W 	imes 1 	imes (20 - 10)$ $8000 = 3000 + 10W$ $10W = 5000 \implies W = 500 \ g$. Step $2$: Heat lost by mixture = Heat gained by metallic block. Mass of mixture = $100 \ g + 300 \ g = 400 \ g$. Heat lost by mixture and calorimeter = $(400 	imes 1 + 500) 	imes (20 - 19) = 900 	imes 1 = 900 \ cal$. Heat gained by metallic block = $m 	imes S_b 	imes \Delta T = 1000 \ g 	imes S_b 	imes (19 - 10) = 9000 	imes S_b$. Equating the two: $9000 	imes S_b = 900$. $S_b = 0.1 \ cal/g^{\circ} C$.

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