(A) The heat lost by the water as it cools from $25^{\circ} C$ to $0^{\circ} C$ is used to melt the ice.Heat lost by water: $Q = m_w s_w \Delta \theta

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(A) The heat lost by the water as it cools from $25^{\circ} C$ to $0^{\circ} C$ is used to melt the ice.Heat lost by water: $Q = m_w s_w \Delta \theta$$Q = 0.2 \, kg \times 4200 \, J \, kg^{-1} \, K^{-1} \times (25 - 0) \, K = 21000 \, J$Heat required to melt $m_{ice}$ grams of ice: $Q = m_{ice} L_f$$21000 \, J = m_{ice} \times 3.4 \times 10^{5} \, J \, kg^{-1}$$m_{ice} = \frac{21000}{3.4 \times 10^{5}} \, kg$$m_{ice} = 0.06176 \, kg = 61.76 \, g$Rounding to the nearest value,the amount of ice that melts is approximately $61.7 \, g$.

Class 11 Physics 1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry Principle of Calorimetry and Water Equivalent

Medium

The specific heat of water is $4200 \, J \, kg^{-1} \, K^{-1}$ and the latent heat of ice is $3.4 \times 10^{5} \, J \, kg^{-1}$. $100 \, g$ of ice at $0^{\circ} C$ is placed in $200 \, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of the water reaches $0^{\circ} C$ is close to (in grams):

JEE MAIN 2020 All JEE MAIN

A
$61.7$
B
$63.8$
C
$69.3$
D
$64.6$

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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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Two liquids $A$ and $B$ are at $32^{\circ}C$ and $24^{\circ}C$. When mixed in equal masses,the temperature of the mixture is found to be $28^{\circ}C$. Their specific heats are in the ratio of:

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$A$ piece of metal of mass $5 \ kg$ and temperature $-50 \ ^\circ C$ is dropped in $20 \ kg$ water at $52 \ ^\circ C$ temperature. In thermal equilibrium,the temperature of water decreases by $2 \ ^\circ C$. Find the specific heat of the metal in $Cal/g \ ^\circ C$.

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When $100\,g$ of a liquid $A$ at $100\,^oC$ is added to $50\,g$ of a liquid $B$ at temperature $75\,^oC$,the temperature of the mixture becomes $90\,^oC$. The temperature of the mixture,if $100\,g$ of liquid $A$ at $100\,^oC$ is added to $50\,g$ of liquid $B$ at $50\,^oC$,will be ........$^oC$

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Water is kept in a thermally insulated container at a temperature of $-10^{\circ}C$. If a small piece of ice is dropped into it, what will be the ratio of the mass of ice formed from the water to the initial mass of water?

$A$ liquid of mass $M$ and specific heat $S$ is at a temperature $2t$. If another liquid of thermal capacity $1.5$ times,at a temperature of $\frac{t}{3}$ is added to it,the resultant temperature will be

Two liquids $A$ and $B$ are at $32^{\circ}C$ and $24^{\circ}C$. When mixed in equal masses,the temperature of the mixture is found to be $28^{\circ}C$. Their specific heats are in the ratio of:

$A$ piece of metal of mass $5 \ kg$ and temperature $-50 \ ^\circ C$ is dropped in $20 \ kg$ water at $52 \ ^\circ C$ temperature. In thermal equilibrium,the temperature of water decreases by $2 \ ^\circ C$. Find the specific heat of the metal in $Cal/g \ ^\circ C$.

When $100\,g$ of a liquid $A$ at $100\,^oC$ is added to $50\,g$ of a liquid $B$ at temperature $75\,^oC$,the temperature of the mixture becomes $90\,^oC$. The temperature of the mixture,if $100\,g$ of liquid $A$ at $100\,^oC$ is added to $50\,g$ of liquid $B$ at $50\,^oC$,will be ........$^oC$

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