A 100.0 mL sample of water is heatedto its Boiling Point. How much heat (in kj) is required to vaporize it? (Assume Density of 1.00 g/mL).. .

The few things we have to know before answering. It takes 2257 KJ to vaporize 1 kg of water. This is the latent heart of vaporization. Latent heat is the heat needed to change the phase of the substance without the increase in temperature.

We first convert 100ml to kg. With the density at 1.0 g/ml, one hundred ml is equivalent to 100 grams or 0.1 kg. We then multiply this to the latent heat of vaporization of water which is 2257KJ/kg. We should get 225.7KJ.

Аноним Аноним · Answer 2 · 2016-07-04T10:40:31+02:00

Аноним Аноним

04-07-2016

The latent heat of vaporization of water is 2258 kJ/kg.
The mass of water sample (m) is determined using the equation:
m=ρ*V ; where ρ is the density and V is the volume
m=1*100=100g or 0.1 kg

The heat in kJ required to vaporize it:
Q=mΔHvap=0.1kg*2258kJ/kg= 225.8 kJ

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