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
1 answer:
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 =
Hence, there was 450.068g of water in the pot.
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