(A) Heat gained by ice = Heat lost by water.Heat gained by ice = (Heat to raise ice from $-10^{\circ}C$ to $0^{\circ}C$) + (Heat to melt ice at $0^{\c

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(A) Heat gained by ice = Heat lost by water.Heat gained by ice = (Heat to raise ice from $-10^{\circ}C$ to $0^{\circ}C$) + (Heat to melt ice at $0^{\circ}C$) + (Heat to raise melted water from $0^{\circ}C$ to $T^{\circ}C$).$Q_{gain} = (10 \times 2100 \times 10) + (10 \times 3.36 \times 10^5) + (10 \times 4200 \times T) = 210000 + 3360000 + 42000T = 3570000 + 42000T$.Heat lost by water = $100 \times 4200 \times (25 - T) = 420000 \times (25 - T) = 10500000 - 420000T$.Equating the two: $3570000 + 42000T = 10500000 - 420000T$.$462000T = 6930000$.$T = 6930000 / 462000 = 15^{\circ}C$.The decrement in temperature is $\Delta T = 25^{\circ}C - 15^{\circ}C = 10^{\circ}C$.

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

Medium

$10 \ kg$ of ice at $-10^{\circ}C$ is added to $100 \ kg$ of water to lower its temperature from $25^{\circ}C.$ Consider no heat exchange to surroundings. The decrement to the temperature of water is . . . . . . $^{\circ}C.$ (specific heat of ice $= 2100 \ J/kg.^{\circ}C$, specific heat of water $= 4200 \ J/kg.^{\circ}C$, latent heat of fusion of ice $= 3.36 \times 10^{5} \ J/kg$)

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$15$
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$11.6$

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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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If $540 \, g$ of ice at $0^{\circ}C$ is mixed with $540 \, g$ of water at $80^{\circ}C$,what will be the final temperature of the mixture in $^{\circ}C$?

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Three bodies of the same material and having masses $m, m$ and $3m$ are at temperatures $40^{\circ} C, 50^{\circ} C$ and $60^{\circ} C$ respectively. If the bodies are brought in thermal contact, the final temperature will be (in $^{\circ} C$)

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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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An iron ball of mass $0.2\,kg$ is heated to $100\,^{\circ}C$ and put into a block of ice at $0\,^{\circ}C.$ If $25\,g$ of ice melts,and the latent heat of fusion of ice is $80\,cal/g,$ find the specific heat of iron in $cal/g\,^{\circ}C.$

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$A$ vessel contains $110\,g$ of water. The heat capacity of the vessel is equivalent to $10\,g$ of water. The initial temperature of the water in the vessel is $10\,^{\circ}C$. If $220\,g$ of hot water at $70\,^{\circ}C$ is poured into the vessel,the final temperature,neglecting radiation loss,will be nearly equal to ........ $^{\circ}C$.

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If $540 \, g$ of ice at $0^{\circ}C$ is mixed with $540 \, g$ of water at $80^{\circ}C$,what will be the final temperature of the mixture in $^{\circ}C$?

Three bodies of the same material and having masses $m, m$ and $3m$ are at temperatures $40^{\circ} C, 50^{\circ} C$ and $60^{\circ} C$ respectively. If the bodies are brought in thermal contact, the final temperature will be (in $^{\circ} C$)

$200 \,g$ 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 $= 80 \,cal \,g^{-1}$).

An iron ball of mass $0.2\,kg$ is heated to $100\,^{\circ}C$ and put into a block of ice at $0\,^{\circ}C.$ If $25\,g$ of ice melts,and the latent heat of fusion of ice is $80\,cal/g,$ find the specific heat of iron in $cal/g\,^{\circ}C.$

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