(C) Heat released by $4 \text{ g}$ of steam to condense into water at $100^{\circ} C$ is $Q_1 = m_s L = 4 \times 540 = 2160 \text{ cal}$.Heat required

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(C) Heat released by $4 \text{ g}$ of steam to condense into water at $100^{\circ} C$ is $Q_1 = m_s L = 4 \times 540 = 2160 \text{ cal}$.Heat required by $20 \text{ g}$ of water to reach $100^{\circ} C$ is $Q_2 = m_w c \Delta T = 20 \times 1 \times (100 - 46) = 20 \times 54 = 1080 \text{ cal}$.Since $Q_1 > Q_2$, the steam will not condense completely.The amount of steam condensed is $m' = \frac{Q_2}{L} = \frac{1080}{540} = 2 \text{ g}$.Final mass of water in the container = Initial mass of water + Mass of condensed steam = $20 \text{ g} + 2 \text{ g} = 22 \text{ g}$.

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

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$4 \text{ g}$ of steam at $100^{\circ} C$ is added to $20 \text{ g}$ of water at $46^{\circ} C$ in a container of negligible mass. Assuming no heat is lost to the surroundings, the mass of water in the container at thermal equilibrium is. (Latent heat of vaporisation $= 540 \text{ cal/g}$, Specific heat of water $= 1 \text{ cal/g}^{\circ} C$):- (in $\text{ g}$)

A
$18$
B
$20$
C
$22$
D
$24$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

$A$ block of mass $2.5 \ kg$ is heated to a temperature of $500^{\circ} C$ and placed on a large ice block. What is the maximum amount of ice (in $kg$) that can melt (approx.)? (Specific heat of the body $= 0.1 \ cal/g^{\circ} C$)

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$200 \,g$ of ice at $-20^{\circ} C$ is mixed with $500 \,g$ of water at $20^{\circ} C$ in an insulating vessel. The final mass of water in the vessel is ........... $g$ (specific heat of ice $= 0.5 \,cal \,g^{-1} {^{\circ}C}^{-1}$,latent heat of fusion of ice $= 80 \,cal \,g^{-1}$).

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$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

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How many grams of ice at $0 \, ^\circ \text{C}$ will be melted by $1 \, \text{g}$ of steam at $100 \, ^\circ \text{C}$? (Latent heat of fusion of ice $L = 80 \, \text{cal/g}$ and latent heat of vaporization of water $L' = 540 \, \text{cal/g}$)

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$1 \,kg$ of ice at $-20^{\circ} C$ is mixed with $2 \,kg$ of water at $90^{\circ} C$. Assuming that there is no loss of energy to the environment,the final temperature of the mixture is ............ $^{\circ} C$. (Assume,latent heat of ice $= 334.4 \,kJ/kg$,specific heat of water and ice are $4.18 \,kJ \,kg^{-1} K^{-1}$ and $2.09 \,kJ \,kg^{-1} K^{-1}$,respectively.)

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$A$ block of mass $2.5 \ kg$ is heated to a temperature of $500^{\circ} C$ and placed on a large ice block. What is the maximum amount of ice (in $kg$) that can melt (approx.)? (Specific heat of the body $= 0.1 \ cal/g^{\circ} C$)

$200 \,g$ of ice at $-20^{\circ} C$ is mixed with $500 \,g$ of water at $20^{\circ} C$ in an insulating vessel. The final mass of water in the vessel is ........... $g$ (specific heat of ice $= 0.5 \,cal \,g^{-1} {^{\circ}C}^{-1}$,latent heat of fusion of ice $= 80 \,cal \,g^{-1}$).

$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

How many grams of ice at $0 \, ^\circ \text{C}$ will be melted by $1 \, \text{g}$ of steam at $100 \, ^\circ \text{C}$? (Latent heat of fusion of ice $L = 80 \, \text{cal/g}$ and latent heat of vaporization of water $L' = 540 \, \text{cal/g}$)

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