When 4 liters of water were heated at 55 ° C, 50 g of kerosene burned out in a kerosene stove. What is the efficiency of a primus?

Given:

V = 4 liters = 4 * 10 ^ -3 m ^ 3 – volume of water;

ro = 1000 kg / m ^ 3 – water density;

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

dT = 55 degrees Celsius – the temperature at which the water was heated;

m1 = 50 grams = 0.05 kilograms is the mass of burned kerosene;

q1 = 40.8 * 10 ^ 6 J / kg is the specific heat of combustion of kerosene.

It is required to determine the efficiency of the primus n.

To heat water, heat is required:

Q = c * m * dT = c * ro * V * dT = 4200 * 1000 * 0.004 * 55 = 924000 Joules.

During the combustion of kerosene, heat was released:

Q1 = q1 * m1 = 40.8 * 10 ^ 6 * 0.05 = 2,040,000 Joules.

Then the efficiency of the primus is equal to:

n = Q / Q1 = 924000/2040000 = 0.45 = 45%.

Answer: The efficiency of the primus is 45%.