(D) Let the final equilibrium temperature be $T$. According to the principle of calorimetry, heat lost by the metal ball = heat gained by the water an

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(D) Let the final equilibrium temperature be $T$. According to the principle of calorimetry, heat lost by the metal ball = heat gained by the water and the vessel. $m_{metal} c_{metal} (T_{initial, metal} - T) = (m_{water} c_{water} + C_{vessel}) (T - T_{initial, water})$ $0.1 \times 400 \times (500 - T) = (0.5 \times 4200 + 800) \times (T - 30)$ $40(500 - T) = (2100 + 800) \times (T - 30)$ $20000 - 40T = 2900(T - 30)$ $20000 - 40T = 2900T - 87000$ $107000 = 2940T$ $T = \frac{107000}{2940} \approx 36.39 \, ^\circ C$ Change in temperature $\Delta T = 36.39 - 30 = 6.39 \, ^\circ C$. Percentage increment = $\frac{\Delta T}{T_{initial}} \times 100 = \frac{6.39}{30} \times 100 \approx 21.3 \%$. Rounding to the nearest given option, the answer is $20 \%$.

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

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$A$ metal ball of mass $0.1 \, kg$ is heated up to $500 \, ^\circ C$ and dropped into a vessel of heat capacity $800 \, J K^{-1}$ containing $0.5 \, kg$ of water. The initial temperature of the water and the vessel is $30 \, ^\circ C$. The approximate percentage increment in the temperature of the water is ........ $\%$. [Specific heat capacities of water and metal are $4200 \, J kg^{-1} K^{-1}$ and $400 \, J kg^{-1} K^{-1}$ respectively.]

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A
$15$
B
$30$
C
$25$
D
$20$

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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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$A$ water heater of power $2000\,W$ is used to heat water. The specific heat capacity of water is $4200\,J\,kg^{-1}\,K^{-1}$. The efficiency of the heater is $70\%$. The time required to heat $2\,kg$ of water from $10^{\circ}C$ to $60^{\circ}C$ is $..........s$. (Assume that the specific heat capacity of water remains constant over the temperature range).

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$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

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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 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........ $^{\circ}C$

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When $0.15 \; kg$ of ice at $0^{\circ} C$ is mixed with $0.30 \; kg$ of water at $50^{\circ} C$ in a container,the resulting temperature is $6.7^{\circ} C$. Calculate the latent heat of fusion of ice. (Given: $s_{\text{water}} = 4186 \; J \; kg^{-1} \; K^{-1}$)

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$2 \ g$ of steam at $100 \ ^\circ C$ is passed into $40 \ g$ of water at $25 \ ^\circ C$. The final temperature of the mixture becomes $54.3 \ ^\circ C$. Find the latent heat of vaporization of steam in $cal/g$.

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$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

When $0.15 \; kg$ of ice at $0^{\circ} C$ is mixed with $0.30 \; kg$ of water at $50^{\circ} C$ in a container,the resulting temperature is $6.7^{\circ} C$. Calculate the latent heat of fusion of ice. (Given: $s_{\text{water}} = 4186 \; J \; kg^{-1} \; K^{-1}$)

$2 \ g$ of steam at $100 \ ^\circ C$ is passed into $40 \ g$ of water at $25 \ ^\circ C$. The final temperature of the mixture becomes $54.3 \ ^\circ C$. Find the latent heat of vaporization of steam in $cal/g$.

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