How much kerosene was burned to heat 5 kg water at 55 degrees? The heater efficiency is 60%.

Given:

m = 5 kilograms is the mass of water;

dT = 55 degrees Celsius – the temperature to which you need to heat the water;

c = 4200 J / (kg * C) – specific heat of water;

N = 60% = 0.6 – heater efficiency;

q = 40.8 * 10 ^ 6 J / kg – specific heat of combustion of kerosene.

It is required to determine the mass of kerosene m1 (kilogram), which must be burned to heat the water.

Let’s find the energy that is needed to heat water:

Q = c * m * dT = 4200 * 5 * 55 = 1155000 Joules = 1.2 * 10 ^ 6 J.

Since the efficiency of the heater is 60%, the total consumed energy must be:

Q / Q1 = N;

Q1 = Q / N = 1.2 * 10 ^ 6 / 0.6 = 2 * 10 ^ 6 Joules.

Then the mass of kerosene will be equal to:

m1 * q = Q1;

m1 = Q1 / q = 2 * 10 ^ 6 / (40.8 * 10 ^ 6) = 2 / 40.8 = 0.05 kilograms.

Answer: It is necessary to burn 0.05 kilograms of kerosene.