(C) According to the principle of calorimetry,heat lost by steam = heat gained by ice.Heat lost by $M$ grams of steam at $100^{\circ} C$ to become wat

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(C) According to the principle of calorimetry,heat lost by steam = heat gained by ice.Heat lost by $M$ grams of steam at $100^{\circ} C$ to become water at $40^{\circ} C$:$Q_{lost} = M \times L_v + M \times c_w \times \Delta T$$Q_{lost} = M \times 540 + M \times 1 \times (100 - 40) = 540M + 60M = 600M$Heat gained by $200 \; g$ of ice at $0^{\circ} C$ to become water at $40^{\circ} C$:$Q_{gained} = m_{ice} \times L_f + m_{ice} \times c_w \times \Delta T$$Q_{gained} = 200 \times 80 + 200 \times 1 \times (40 - 0) = 16000 + 8000 = 24000 \; cal$Equating the two:$600M = 24000$$M = 24000 / 600 = 40 \; g$.

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

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$M$ grams of steam at $100^{\circ} C$ is mixed with $200 \; g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [heat of vaporization of water is $540 \; cal/g$ and heat of fusion of ice is $80 \; cal/g$],the value of $M$ is:

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A
$35$
B
$37$
C
$40$
D
$42$

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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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$M$ grams of steam at $100^{\circ} C$ is mixed with $200 \; g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [heat of vaporization of water is $540 \; cal/g$ and heat of fusion of ice is $80 \; cal/g$],the value of $M$ is:

Similar Questions

$200 \, g$ of a solid ball at $20 \, ^\circ C$ is dropped in an equal amount of water at $80 \, ^\circ C$. The resulting temperature is $60 \, ^\circ C$. This means that the specific heat of the solid is

$A$ 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 $Cal/g \ ^\circ C$.

$A$ liquid of mass $m$ and specific heat $c$ is at a temperature of $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is at a temperature of $T$. What will be the final temperature of the mixture?

$37 \ g$ of ice at $0^{\circ} C$ temperature is mixed with $74 \ g$ of water at $70^{\circ} C$ temperature. The resultant temperature is (Specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$) (in $^{\circ} C$)

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)

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