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If one kilogram water at 100^(@)C is vap...

If one kilogram water at `100^(@)C` is vapourised in open atmosphere. The correct statement is

A

increase in internal energy is equal to L (L is latent heat of vaporisation for 1 Kg)

B

increase in internal energy is zero

C

increase in internal energy is less than L

D

none of the above

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the process of vaporization of water at 100°C and determine the correct statement regarding the increase in internal energy. ### Step-by-Step Solution: 1. **Identify Given Data:** - Mass of water, \( m = 1 \, \text{kg} \) - Temperature of water, \( T = 100^\circ C \) - Latent heat of vaporization of water, \( L \) (approximately \( 2260 \, \text{kJ/kg} \) or \( 2.26 \times 10^6 \, \text{J/kg} \)) 2. **Calculate Heat Supplied (Q):** - The heat supplied to vaporize the water can be calculated using the formula: \[ Q = m \times L \] - Substituting the values: \[ Q = 1 \, \text{kg} \times L \] - Thus, \( Q = L \). 3. **Understand the Work Done (W):** - When water vaporizes, it expands against the atmospheric pressure. The work done during this expansion is given by: \[ W = P \Delta V \] - Here, \( P \) is the atmospheric pressure and \( \Delta V \) is the change in volume when water turns into steam. 4. **Apply the First Law of Thermodynamics:** - According to the first law of thermodynamics: \[ \Delta U = Q - W \] - Where \( \Delta U \) is the change in internal energy. 5. **Substituting Values:** - Since \( Q = L \), we can substitute this into the equation: \[ \Delta U = L - W \] 6. **Analyze the Relationship:** - Since the work done \( W \) is positive (because the system does work on the surroundings during expansion), it follows that: \[ \Delta U < L \] - This means the increase in internal energy is less than the latent heat of vaporization. 7. **Conclusion:** - The correct statement regarding the increase in internal energy when 1 kg of water at \( 100^\circ C \) is vaporized in open atmosphere is that the increase in internal energy is less than the latent heat of vaporization. ### Final Answer: The correct statement is: The increase in internal energy is less than the latent heat of vaporization. ---

To solve the problem, we need to analyze the process of vaporization of water at 100°C and determine the correct statement regarding the increase in internal energy. ### Step-by-Step Solution: 1. **Identify Given Data:** - Mass of water, \( m = 1 \, \text{kg} \) - Temperature of water, \( T = 100^\circ C \) - Latent heat of vaporization of water, \( L \) (approximately \( 2260 \, \text{kJ/kg} \) or \( 2.26 \times 10^6 \, \text{J/kg} \)) ...
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