Question

Calculate the amount of heat required to vaporize 84.8 g of water at its boiling point.

Answer 1

You need to use the specifc latent heat of vaporization of water which you can find in internet.

I found 2,264.76 KJ / kg = 2,264.76 J/g

Now use total heat = specific latent heat * mass

=> 84.8 g * 2,264.76 J/ g = 192,051.65 J.

Answer: 192,051.65 J

Answer 2

Answer: Amount of heat required to vaporize 84.8 g of water is 191.7kJ.

Explanation: We are given 84.8 grams of water, to convert it into moles, we use the formula:

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of water = 18 g/mol

`Moles=(84.8g)/(18g/mol)=4.71moles`

As we know that
`\Delta H_(vap)` for 1 mole of water at 100° C is 40.7 kJ/mol

So, to calculate the amount of heat required, we use the formula:

`q=m\Delta H_(vap)`

Putting the values in above equation, we get

`q=4.71mol* 40.7kJ/mol`

q = 191.7kJ