Answer:
13598 J
Step-by-step 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.