(A) Let $m$ be the mass of ice that melts to reach equilibrium at $0\,^{\circ}C$.Heat lost by water $= m_w \cdot c_w \cdot \Delta T = 100 \times 1 \ti

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(A) Let $m$ be the mass of ice that melts to reach equilibrium at $0\,^{\circ}C$.Heat lost by water $= m_w \cdot c_w \cdot \Delta T = 100 \times 1 \times (30 - 0) = 3000\, cal$.Heat gained by ice to melt $= m \cdot L_f = m \times 80$.Equating heat lost and heat gained: $3000 = 80m$.Solving for $m$: $m = \frac{3000}{80} = 37.5\, g$.This is the amount of ice that melts.The initial mass of ice was $50\, g$.Therefore,the amount of ice left $= 50 - 37.5 = 12.5\, g$.

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

Medium

$50\, g$ of ice at $0\,^{\circ}C$ is dropped into a calorimeter containing $100\, g$ of water at $30\,^{\circ}C$. If the thermal capacity of the calorimeter is zero,then the amount of ice left in the mixture at equilibrium is ........ $g$.

A
$12.5$
B
$25$
C
$20$
D
$10$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

$80 \ gm$ of water at $30 \ ^\circ C$ is poured onto a large block of ice at $0 \ ^\circ C$. The mass of ice that melts is .... $gm$

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$50 \, g$ of ice at $0^\circ C$ is placed in an insulated vessel. $50 \, g$ of water at $100^\circ C$ is mixed into it. Neglecting heat loss,what is the final temperature of the mixture?

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$100 \, g$ of ice at $0^{\circ}C$ is mixed with $100 \, g$ of water at $100^{\circ}C$. What will be the final temperature of the mixture in $^{\circ}C$?

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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 $cal/g$?

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When $x \text{ grams}$ of steam at $100^{\circ}C$ is mixed with $y \text{ grams}$ of ice at $0^{\circ}C$,we obtain $(x + y) \text{ grams}$ of water at $100^{\circ}C$. What is the ratio $y/x$?

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$50\, g$ of ice at $0\,^{\circ}C$ is dropped into a calorimeter containing $100\, g$ of water at $30\,^{\circ}C$. If the thermal capacity of the calorimeter is zero,then the amount of ice left in the mixture at equilibrium is ........ $g$.

Similar Questions

$80 \ gm$ of water at $30 \ ^\circ C$ is poured onto a large block of ice at $0 \ ^\circ C$. The mass of ice that melts is .... $gm$

$50 \, g$ of ice at $0^\circ C$ is placed in an insulated vessel. $50 \, g$ of water at $100^\circ C$ is mixed into it. Neglecting heat loss,what is the final temperature of the mixture?

$100 \, g$ of ice at $0^{\circ}C$ is mixed with $100 \, g$ of water at $100^{\circ}C$. What will be the final temperature of the mixture in $^{\circ}C$?

$2 \; g$ 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 $cal/g$?

When $x \text{ grams}$ of steam at $100^{\circ}C$ is mixed with $y \text{ grams}$ of ice at $0^{\circ}C$,we obtain $(x + y) \text{ grams}$ of water at $100^{\circ}C$. What is the ratio $y/x$?

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