What amount of water with a specific heat of 4.2 * 10 ^ 3 J / kg * 0C can be heated by 20 0C, spending 29.4 MJ of heat?

Initial data: Δt (specified change in water temperature) = 20 ºС; Q (consumed amount of heat for heating water) = 29.4 MJ (29.4 * 10 ^ 6 J).
Reference values: according to the condition Sv (specific heat capacity of water) = 4.2 * 10 ^ 3 J / (kg * ºС).
The amount of heated water (mass of water) can be expressed from the formula: Q = Cw * m * Δt, whence m = Q / (Cw * Δt).
Calculation: m = 29.4 * 10 ^ 6 / (4.2 * 10 ^ 3 * 20) = 350 kg.
Answer: 350 kg of water can be heated by 20 ºС.