How much water can be heated from 20 ° C to 100 ° C, while consuming 672 kJ of heat?

Problem data: t1 (initial temperature of the required water mass) = 20 ºС; t2 (final temperature) = 100 ºС; Q (heat consumption) = 672 kJ (672 * 10 ^ 3 J).

Reference values: Sv (specific heat capacity of water) = 4200 J / (kg * K).

We express the sought mass of heated water from the formula: Q = Cw * m * (t2 – t1) and m = Q / (Cw * (t2 – t1)).

Let’s calculate: m = 672 * 10 ^ 3 / (4200 * (100 – 20)) = 2 kg.

Answer: Due to the consumption of 672 kJ of heat, it will be possible to heat 2 kilograms of water.