Question:
The heat extracted out of \( x \) gram of water initially at \( 50^\circ\text{C} \) to cool it down to \( 0^\circ\text{C} \) is sufficient to evaporate \( (1000-x) \) gram of water also initially at \( 50^\circ\text{C} \). The value of \( x \) (closest integer) is \dots
(Take latent heat of water 2256 kJ/kg, specific heat capacity of water 4200 J/kg$\cdot$K)}
The heat extracted out of \( x \) gram of water initially at \( 50^\circ\text{C} \) to cool it down to \( 0^\circ\text{C} \) is sufficient to evaporate \( (1000-x) \) gram of water also initially at \( 50^\circ\text{C} \). The value of \( x \) (closest integer) is \dots
(Take latent heat of water 2256 kJ/kg, specific heat capacity of water 4200 J/kg$\cdot$K)}
(Take latent heat of water 2256 kJ/kg, specific heat capacity of water 4200 J/kg$\cdot$K)}
Updated On: Apr 12, 2026
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Correct Answer: 922
Solution and Explanation
Step 1: Understanding the Concept:
This problem involves the principle of calorimetry where heat lost by one substance equals heat gained by another. One part of the water loses sensible heat during cooling. The other part gains sensible heat to reach boiling point and then gains latent heat to evaporate.
Step 2: Detailed Explanation:
Let \( s = 4200 \text{ J/kg}\cdot\text{K} \) and \( L = 2256 \times 10^3 \text{ J/kg} \).
Heat Lost by \( x \) grams of water:
Cooling from \( 50^\circ\text{C} \) to \( 0^\circ\text{C} \):
\[ Q_{\text{lost}} = m \cdot s \cdot \Delta T = \left(\frac{x}{1000}\right) \cdot 4200 \cdot (50 - 0) = 210x \text{ J} \].
Heat Gained by \( (1000-x) \) grams of water:
1. Heating from \( 50^\circ\text{C} \) to \( 100^\circ\text{C} \):
\[ Q_1 = m \cdot s \cdot \Delta T' = \left(\frac{1000-x}{1000}\right) \cdot 4200 \cdot (100 - 50) = 210(1000-x) \text{ J} \].
2. Evaporation at \( 100^\circ\text{C} \):
\[ Q_2 = m \cdot L = \left(\frac{1000-x}{1000}\right) \cdot 2256 \times 10^3 = 2256(1000-x) \text{ J} \].
Total heat gained:
\[ Q_{\text{gained}} = (210 + 2256)(1000-x) = 2466(1000-x) \text{ J} \].
Equating heat lost and heat gained:
\[ 210x = 2466(1000-x) \]
\[ 210x = 2466000 - 2466x \implies 2676x = 2466000 \]
\[ x = \frac{2466000}{2676} \approx 921.52 \].
Step 3: Final Answer:
The value of \( x \) to the closest integer is 922.
This problem involves the principle of calorimetry where heat lost by one substance equals heat gained by another. One part of the water loses sensible heat during cooling. The other part gains sensible heat to reach boiling point and then gains latent heat to evaporate.
Step 2: Detailed Explanation:
Let \( s = 4200 \text{ J/kg}\cdot\text{K} \) and \( L = 2256 \times 10^3 \text{ J/kg} \).
Heat Lost by \( x \) grams of water:
Cooling from \( 50^\circ\text{C} \) to \( 0^\circ\text{C} \):
\[ Q_{\text{lost}} = m \cdot s \cdot \Delta T = \left(\frac{x}{1000}\right) \cdot 4200 \cdot (50 - 0) = 210x \text{ J} \].
Heat Gained by \( (1000-x) \) grams of water:
1. Heating from \( 50^\circ\text{C} \) to \( 100^\circ\text{C} \):
\[ Q_1 = m \cdot s \cdot \Delta T' = \left(\frac{1000-x}{1000}\right) \cdot 4200 \cdot (100 - 50) = 210(1000-x) \text{ J} \].
2. Evaporation at \( 100^\circ\text{C} \):
\[ Q_2 = m \cdot L = \left(\frac{1000-x}{1000}\right) \cdot 2256 \times 10^3 = 2256(1000-x) \text{ J} \].
Total heat gained:
\[ Q_{\text{gained}} = (210 + 2256)(1000-x) = 2466(1000-x) \text{ J} \].
Equating heat lost and heat gained:
\[ 210x = 2466(1000-x) \]
\[ 210x = 2466000 - 2466x \implies 2676x = 2466000 \]
\[ x = \frac{2466000}{2676} \approx 921.52 \].
Step 3: Final Answer:
The value of \( x \) to the closest integer is 922.
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