how much heat is absorbed when one kg of a liquid gets vaporised at its boiling point?

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Solution

we know that
Q = mL

here
Q is the heat absorbed
m is the mass of the substance (water) = 1 kg
l is the latent heap of vaporization (of water) = 2257 KJ/kg

thus,
Q = 1 x 2257
so, we get

Q = 2257 KJ of heat absorbed

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how much heat is absorbed when one kg of a liquid gets vaporised at its boiling point?

we know thatQ = mL hereQ is the heat absorbedm is the mass of the substance (water) = 1 kgl is the latent heap of vaporization (of water) = 2257 KJ/kg thus,Q = 1 x 2257so, we get Q = 2257 KJ of heat absorbed

we know that
Q = mL

here
Q is the heat absorbed
m is the mass of the substance (water) = 1 kg
l is the latent heap of vaporization (of water) = 2257 KJ/kg

thus,
Q = 1 x 2257
so, we get

Q = 2257 KJ of heat absorbed