Will burns a 0.3 g nut beneath 52 g of water, which increases in temperature from 22°C to 50°C.?

Question

Will burns a 0.3 g nut beneath 52 g of water, which increases in temperature from 22°C to 50°C. (a) Assuming 40% efficiency, what is the food value in calories of the nut? (in calories)

Answer 1

(50 - 22) = 28° change
It takes 1 calorie to increase the temperature of 1 g of water by 1°C, so:
(52)(28) = 1456 calories

Since we assume 40% efficiency, the food value is:
(.4)(1456) = 582.4 calories

Answer 2

I had a different interpretation than Steve. It could be that the burning of the nut and heating of the water was 40% efficient. Meaning some of the heat excaped and did not heat the water. But if you eat it, you get all the calories into your system. I don't know which is right.

If I'm right, it would be
40% of food value = 1456 calories
food value = 1456 calories/0.4 = 3640 calories

It's your call which interpretation is right.