(C) Using the principle of calorimetry,heat lost by liquid $A$ = heat gained by liquid $B$.$m_A S_A (T_{A,initial} - T_{mix}) = m_B S_B (T_{mix} - T_{

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(C) Using the principle of calorimetry,heat lost by liquid $A$ = heat gained by liquid $B$.$m_A S_A (T_{A,initial} - T_{mix}) = m_B S_B (T_{mix} - T_{B,initial})$$100 \times S_A \times (100 - 90) = 50 \times S_B \times (90 - 75)$$1000 S_A = 750 S_B$$S_A = 0.75 S_B = \frac{3}{4} S_B$Now,for the second case,let the final temperature be $T$:$100 \times S_A \times (100 - T) = 50 \times S_B \times (T - 50)$Substitute $S_A = \frac{3}{4} S_B$:$100 \times (\frac{3}{4} S_B) \times (100 - T) = 50 \times S_B \times (T - 50)$$75 (100 - T) = 50 (T - 50)$$3 (100 - T) = 2 (T - 50)$$300 - 3T = 2T - 100$$5T = 400$$T = 80\,^oC$

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

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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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A
$85$
B
$60$
C
$80$
D
$70$

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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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$A$ piece of metal of $850 \text{ K}$ is dropped into $1 \text{ kg}$ of water at $300 \text{ K}$. If the equilibrium temperature of the mixture is $350 \text{ K}$, then the heat capacity of the metal expressed in $\text{J/K}$ is (Specific heat of water = $4200 \text{ J/kg} \cdot \text{K}$)

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In a waterfall,the water falls from a height of $100 \ m$. If the entire kinetic energy $(K.E.)$ of the water is converted into heat,the rise in temperature of the water will be ........ $^\circ C$.

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One kilogram of ice at $0\,^{\circ}C$ is mixed with one kilogram of water at $80\,^{\circ}C$. The temperature of the mixture is ........ $^{\circ}C$ (specific heat of water $= 4200\,J\,kg^{-1}\,K^{-1}$ and latent heat of ice $= 336\,kJ\,kg^{-1}$).

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How many grams of a liquid with a specific heat of $0.2 \, cal/g^{\circ}C$ at a temperature of $40^{\circ}C$ must be mixed with $100 \, g$ of a liquid with a specific heat of $0.5 \, cal/g^{\circ}C$ at a temperature of $20^{\circ}C$,so that the final temperature of the mixture becomes $32^{\circ}C$?

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The water equivalent of $20 \,g$ of aluminium (specific heat $0.2 \,cal \,g^{-1} {}^{\circ}C^{-1}$) is ......... $g$.

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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 ........$^oC$

Similar Questions

$A$ piece of metal of $850 \text{ K}$ is dropped into $1 \text{ kg}$ of water at $300 \text{ K}$. If the equilibrium temperature of the mixture is $350 \text{ K}$, then the heat capacity of the metal expressed in $\text{J/K}$ is (Specific heat of water = $4200 \text{ J/kg} \cdot \text{K}$)

In a waterfall,the water falls from a height of $100 \ m$. If the entire kinetic energy $(K.E.)$ of the water is converted into heat,the rise in temperature of the water will be ........ $^\circ C$.

One kilogram of ice at $0\,^{\circ}C$ is mixed with one kilogram of water at $80\,^{\circ}C$. The temperature of the mixture is ........ $^{\circ}C$ (specific heat of water $= 4200\,J\,kg^{-1}\,K^{-1}$ and latent heat of ice $= 336\,kJ\,kg^{-1}$).

The water equivalent of $20 \,g$ of aluminium (specific heat $0.2 \,cal \,g^{-1} {}^{\circ}C^{-1}$) is ......... $g$.

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