(D) Let $m$ be the mass of ice that melts to reach a final equilibrium temperature of $0^{\circ} C$.According to the principle of calorimetry,Heat los

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(D) Let $m$ be the mass of ice that melts to reach a final equilibrium temperature of $0^{\circ} C$.According to the principle of calorimetry,Heat lost by water = Heat gained by ice.Heat lost by $100 \,g$ of water cooling from $80^{\circ} C$ to $0^{\circ} C$ is $Q_{lost} = m_w \cdot s_w \cdot \Delta T = 100 \times 1 \times (80 - 0) = 8000 \,cal$.Heat required to melt $m$ grams of ice at $0^{\circ} C$ is $Q_{gained} = m \cdot L_f = m \times 80$.Equating the two: $80m = 8000$.Solving for $m$: $m = 100 \,g$.Since the total mass of ice is $150 \,g$,the amount of ice that does not melt is $150 \,g - 100 \,g = 50 \,g$.

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

Medium

$150 \,g$ of ice is mixed with $100 \,g$ of water at temperature $80^{\circ} C$. The latent heat of ice is $80 \,cal/g$ and the specific heat of water is $1 \,cal/g^{\circ} C$. Assuming no heat loss to the environment,the amount of ice which does not melt is ........... $g$.

KVPY 2012 All KVPY

A
$100$
B
$0$
C
$150$
D
$50$

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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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$A$ copper block of mass $5.0 \, kg$ is heated to a temperature of $500^{\circ} C$ and is placed on a large ice block. What is the maximum amount of ice (in $kg$) that can melt? [Specific heat of copper: $0.39 \, J g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of water: $335 \, J g^{-1}$]

MediumJEE MAIN 2022

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$2 \; g$ of steam condenses when passed through $40 \; g$ of water initially at $25 \; ^\circ C$. The condensation of steam raises the temperature of water to $54.3 \; ^\circ C$. What is the latent heat of steam in $cal/g$?

Medium

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At a power station,heat is removed from the heat exchanger by cooling water at $6.7 \times 10^9 \ J$ per minute. The cooling water enters at $6.0 \ ^oC$ and leaves at $14.0 \ ^oC$. [Take the specific heat capacity of water $4200 \ J/kg \ ^oC$]. Which of the following is the rate of water flow?

Medium

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Steam is passed into $22 \, g$ of water at $20^{\circ}C$. The mass of water that will be present when the water acquires a temperature of $90^{\circ}C$ is ........ $g$. (Latent heat of steam is $540 \, cal/g$)

Difficult

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$10\; g$ of ice cubes at $0\;^{\circ} C$ are released in a tumbler (water equivalent $55\; g$) at $40\;^{\circ} C$. Assuming that negligible heat is taken from the surroundings,the temperature (in $^{\circ} C$) of water in the tumbler becomes nearly $(L_f = 80\; cal/g)$.

MediumAIPMT 1988

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$150 \,g$ of ice is mixed with $100 \,g$ of water at temperature $80^{\circ} C$. The latent heat of ice is $80 \,cal/g$ and the specific heat of water is $1 \,cal/g^{\circ} C$. Assuming no heat loss to the environment,the amount of ice which does not melt is ........... $g$.

Similar Questions

$A$ copper block of mass $5.0 \, kg$ is heated to a temperature of $500^{\circ} C$ and is placed on a large ice block. What is the maximum amount of ice (in $kg$) that can melt? [Specific heat of copper: $0.39 \, J g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of water: $335 \, J g^{-1}$]

$2 \; g$ of steam condenses when passed through $40 \; g$ of water initially at $25 \; ^\circ C$. The condensation of steam raises the temperature of water to $54.3 \; ^\circ C$. What is the latent heat of steam in $cal/g$?

At a power station,heat is removed from the heat exchanger by cooling water at $6.7 \times 10^9 \ J$ per minute. The cooling water enters at $6.0 \ ^oC$ and leaves at $14.0 \ ^oC$. [Take the specific heat capacity of water $4200 \ J/kg \ ^oC$]. Which of the following is the rate of water flow?

Steam is passed into $22 \, g$ of water at $20^{\circ}C$. The mass of water that will be present when the water acquires a temperature of $90^{\circ}C$ is ........ $g$. (Latent heat of steam is $540 \, cal/g$)

$10\; g$ of ice cubes at $0\;^{\circ} C$ are released in a tumbler (water equivalent $55\; g$) at $40\;^{\circ} C$. Assuming that negligible heat is taken from the surroundings,the temperature (in $^{\circ} C$) of water in the tumbler becomes nearly $(L_f = 80\; cal/g)$.

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