Question

Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of water were in the pot?

Answer 1

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = $2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x$

$2330.35x=1048815\\x=450.068g$

Hence, there was 450.068g of water in the pot.