How much heat is required to convert 1 kg of water at 100°C to steam at 100°C? (

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Question: How much heat is required to convert 1 kg of water at 100°C to steam at 100°C? (Latent heat of vaporization = 2260 J/g)

Options:

Correct Answer: 2260000 J

Solution:

Q = m * L = 1000 g * 2260 J/g = 2260000 J.

How much heat is required to convert 1 kg of water at 100°C to steam at 100°C? (Latent heat of vaporization = 2260 J/g)

How much heat is required to convert 1 kg of water at 100°C to steam at 100°C? (Latent heat of vaporization = 2260 J/g)

Step 1: Identify the mass of water you want to convert to steam. In this case, it is 1 kg.
Step 2: Convert the mass from kilograms to grams because the latent heat of vaporization is given in J/g. 1 kg = 1000 g.
Step 3: Identify the latent heat of vaporization, which is given as 2260 J/g.
Step 4: Use the formula Q = m * L, where Q is the heat required, m is the mass in grams, and L is the latent heat of vaporization.
Step 5: Substitute the values into the formula: Q = 1000 g * 2260 J/g.
Step 6: Calculate the result: Q = 2260000 J.

Latent Heat of Vaporization – The amount of heat required to convert a unit mass of a substance from liquid to gas at constant temperature.
Mass Conversion – Understanding the conversion of mass units from kilograms to grams for calculations.
Heat Calculation – Applying the formula Q = m * L to calculate the heat required for phase changes.