(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 1 Thermal Properties of Matter: Thermometry, 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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1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

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An electric kettle (rated at $2.5\, kW$) is used to heat $3\, kg$ of water from $15\, ^oC$ to its boiling point. It takes $9.5$ minutes. The amount of heat lost is:

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Medium

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Ice at $0^oC$ is added to $200\,g$ of water initially at $70^oC$ in a vacuum flask. When $50\,g$ of ice has been added and has all melted,the temperature of the flask and contents is $40^oC$. When a further $80\,g$ of ice has been added and has all melted,the temperature of the whole is $10^oC$. Calculate the specific latent heat of fusion of ice. [Take $S_w = 1\,cal/g^oC$.]

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An electric kettle (rated at $2.5\, kW$) is used to heat $3\, kg$ of water from $15\, ^oC$ to its boiling point. It takes $9.5$ minutes. The amount of heat lost is:

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)

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