Answer : Heat required will be 76.42 KJ
Explanation : Here we have the data as 33.8 g of water and temperature is 100°C
To calculate the heat required we need to use the formula as,
q= mol of water / heat of vaporization of water.
So we have convert the mass into moles,
33.8g / 18g = 1.877 moles of water.
and heat of vaporization is 40.7 KJ
Now, q = 1.877 X 40.7 KJ = 76.4 KJ
So the heat required will be 76.4 KJ
