What mass of water can be heated from 10 to 100 C, spending 1.5 kW x 1 h of energy

What mass of water can be heated from 10 to 100 C, spending 1.5 kW x 1 h of energy, if only 80% of the energy goes to heating the liquid?

Given: t1 = 10 oC, t2 = 100 oC, E = 1500 W * h, n = 0.8. Find m.

The amount of heat required for heating is determined by the formula Q = C * m * (t2 – t1), where C is the heat capacity of water, J / (kg * oC); m is the mass of water, kg; t1 – initial water temperature, oC; t2 – final water temperature, oC. Hence m = Q / (C * (t2 – t1)). The useful amount of heat Q is defined as Q = E * n, where n is the heating efficiency. We convert watt-hours into joules E = 1500 W * h = 1500 * 3600 J = 5400000 J. According to the reference book, we find C = 4180 J / (kg * oC). Substituting the data into the formulas, we consider: Q = E * n = 5,400,000 * 0.8 = 4,320,000 J; m = Q / (C * (t2 – t1)) = 4320000 / (4180 * (100 – 10)) = 11.5 kg.

Answer: m = 11.5 kg