Abackpacker wants to carry enough fuel to heat 2.3 kg of water from 29 ∘c to 100.0 ∘c. part a if the fuel he carries produces 36 kj of heat per gram when it burns, how much fuel should he carry? (for the sake of simplicity, assume that the transfer of heat is 100 % efficient.)

Answer is: he should carry 18,98 g of fuel.
m(H₂O) = 2,3 kg = 2300 g.
ΔT(H₂O) = 100°C - 29°C = 71°C (71K).
cp(H₂O) = 4,184 J/g·K, specific heat capacity of water.
Q = m(H₂O) · ΔT(H₂O) · cp(H₂O).
Q = 2300 g ·71 K · 4,184 J/g·K.
Q = 683247,2 J = 683,25 kJ.
m(fuel) = 683,25 kJ ÷ 36 kJ/g.
m(fuel) = 18,98 g.
Q - absorbed energy.

North atlantic current or north atlantic drift is the answer