(B) When both boxes are brought into thermal contact,the heat lost by the helium gas equals the heat gained by the nitrogen gas.Heat lost by Helium =

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(B) When both boxes are brought into thermal contact,the heat lost by the helium gas equals the heat gained by the nitrogen gas.Heat lost by Helium = Heat gained by Nitrogen$\mu_B (C_V)_{He} \left( \frac{7}{3} T_0 - T_f \right) = \mu_A (C_V)_{N_2} (T_f - T_0)$Given $\mu_A = 1, \mu_B = 1$,$(C_V)_{He} = \frac{3}{2}R$,and $(C_V)_{N_2} = \frac{5}{2}R$ (as $N_2$ is diatomic and $He$ is monatomic).Substituting the values:$1 \cdot \frac{3}{2}R \left( \frac{7}{3} T_0 - T_f \right) = 1 \cdot \frac{5}{2}R (T_f - T_0)$$3 \left( \frac{7}{3} T_0 - T_f \right) = 5 (T_f - T_0)$$7 T_0 - 3 T_f = 5 T_f - 5 T_0$$12 T_0 = 8 T_f$$T_f = \frac{12}{8} T_0 = \frac{3}{2} T_0$

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

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Two different rigid boxes are placed on a table,each containing a different gas. Box $A$ contains $1 \text{ mole}$ of nitrogen gas at temperature $T_0$,and box $B$ contains $1 \text{ mole}$ of helium gas at temperature $(7/3) T_0$. If these two boxes are brought into thermal contact,heat exchange occurs until a final equilibrium temperature $T_f$ is reached. The final temperature $T_f$ is:

A
$(7/3) T_0$
B
$(3/2) T_0$
C
$(5/2) T_0$
D
$(3/7) T_0$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Principle of Calorimetry and Water Equivalent

The temperature of water of mass $100 \ g$ is raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam to it. The mass of the steam added is (Latent heat of steam $= 540 \ cal \ g^{-1}$ and specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

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$A$ 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?

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$1\, g$ of steam at $100^{\circ}C$ melts ........ $g$ of ice at $0^{\circ}C?$ (Latent heat of ice $= 80\, cal/g$ and latent heat of steam $= 540\, cal/g$)

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$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is:

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$A$ piece of ice (heat capacity $= 2100 \ J \ kg^{-1} \ ^{\circ}C^{-1}$ and latent heat $= 3.36 \times 10^5 \ J \ kg^{-1}$) of mass $m$ grams is at $-5^{\circ}C$ at atmospheric pressure. It is given $420 \ J$ of heat so that the ice starts melting. Finally,when the ice-water mixture is in equilibrium,it is found that $1 \ g$ of ice has melted. Assuming there is no other heat exchange in the process,the value of $m$ is:

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The temperature of water of mass $100 \ g$ is raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam to it. The mass of the steam added is (Latent heat of steam $= 540 \ cal \ g^{-1}$ and specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

$A$ 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?

$1\, g$ of steam at $100^{\circ}C$ melts ........ $g$ of ice at $0^{\circ}C?$ (Latent heat of ice $= 80\, cal/g$ and latent heat of steam $= 540\, cal/g$)

$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is:

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