How much energy is required to boil 65 grams of 100°C water
And then heat the steam to 150°C?

13598 J

Explanation:

Q = m × c × ∆T

Where;

Q = amount of energy (J)

m = mass (grams)

c = specific heat capacity

∆T = change in temperature

m = 65g, specific heat capacity of water = 4.184J/g°C, initial temperature= 100°C, final temperature = 150°C

Q = 65 × 4.184 × (150 - 100)

Q = 271.96 × 50

Q = 13598 J

Hence, 13598 J of energy is required to boil 65 grams of 100°C water and then heat the steam to 150°C.

the answer is b. dilution