(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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(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. $100 \times 1 \times (100 - 20) = 300 \times 1 \times (20 - 10) + W \times 1 \times (20 - 10)$ $8000 = 3000 + 10W$ $10W = 5000 \implies W = 500 \ g$. 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 = $(400 \times 1 + 500) \times (20 - 19) = 900 \times 1 = 900 \ cal$. Heat gained by metallic block = $m \times S_b \times \Delta T = 1000 \ g \times S_b \times (19 - 10) = 9000 \times S_b$. Equating the two: $9000 \times S_b = 900$. $S_b = 0.1 \ cal/g^{\circ} C$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Principle of Calorimetry and Water Equivalent

Medium

When $100 \ g$ of boiling water at $100^{\circ} C$ is added into a calorimeter containing $300 \ g$ of cold water at $10^{\circ} C$, the temperature of the mixture becomes $20^{\circ} C$. Then, a metallic block of mass $1 \ kg$ at $10^{\circ} C$ is dipped into the mixture in the calorimeter. After reaching thermal equilibrium, the final temperature becomes $19^{\circ} C$. What is the specific heat of the metal in $C$.$G$.$S$. units?

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A
$0.01$
B
$0.3$
C
$0.09$
D
$0.1$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

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