A heater marked 60w evaporates 6×10^-3kg of boiling water in 60secs .what is the specific latent heat of vaporization of Water in jkg^-1?
2 answers
The specific latent heat of vaporization of water is 2256 J/kg.
You mean 2256 kJ/kg I think.
but using the data provided:
60 watts * 60 seconds = 3600 Joules input
3600 Joules / 0.006 kg = 600,000 Joules/kg = 600 kJ/kg
I think not
but using the data provided:
60 watts * 60 seconds = 3600 Joules input
3600 Joules / 0.006 kg = 600,000 Joules/kg = 600 kJ/kg
I think not