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1 year ago

a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1

Physics

1 answer:

geniusboy [140]1 year ago

4 0

Your question doesn't look right, so lemme assume you meant this.

A heater marked 60w evaporate 6x 10^-³kg of boiling water in 60 seconds. What is the specific latent heat of vaporization of water in jkg?

so you can understand it better..

Latent heat of evaporation is the heat required required to change water to vapor at the same temperature.(100°C)

eg. water boils at 100°C and in presence of more heat it turns to vapor at that Same temperature.

This heat is know as latent heat of Vaporization.

it's give by H=mL

where L is the specific latent heat of vaporization.

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a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1​

a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1