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%.