I believe the answer is the second one.
Answer:
12552 J or 3000 calories
Explanation:
Q = m × c × ∆T
Where;
Q = amount of heat energy (J)
m = mass of water (g)
c = specific heat capacity (4.184 J/g°C)
∆T = change in temperature
For 50mL of water, there are 50g, hence, m = 50g, c = 4.184 J/g°C, initial temperature = 0°C, final temperature = 60°C.
Q = m × c × ∆T
Q = 50 × 4.184 × (60 - 0)
Q = 209.2 × 60
Q = 12552 J
Hence, the amount of heat energy used to heat the water is 12552 J or 3000 calories
Efficiency = Power Output / Power Input
Power Input = Rate of Energy input = 44.4 MJ/kg * 5 kg/h
= 222 MJ/h
But 1 hour = 3600seconds
222 MJ/h = 222 MJ/3600s = 0.061667 MW J/s = Watts
Power input = 0.061667 MW = 61 667 W
From Efficiency = Power Output / Power Input
28% = Power Output / 61667
Power Output = 0.28 * 61667
Power Output = 17266.76 W
Power Output ≈ 17 267 W
Rate of heat rejection = Power Input - Power Output
= 61667 - 17267 = 44400 W
Rate of heat rejection = 44 400W.
