On a kerosene burner, water with a mass of m1 = 1.2 kg was heated from a temperature of 18 C to boiling.

On a kerosene burner, water with a mass of m1 = 1.2 kg was heated from a temperature of 18 C to boiling. Determine the mass of burned kerosene if the loss to the environment is 40%. (Specific heat of combustion of kerosene q = 43000000 J / kg, specific heat of water c = 4200 J / kg * C, boiling point of water 100 C).

m1 = 1.2 kg

t1 = 18 ° C.

t2 = 100 ° C.

q = 43,000,000 J / kg.

C = 4200 J / kg * ° С.

Efficiency = 60%.

m2 -?

Since the losses are 40%, only 100% – 40% = 60% of the thermal energy obtained during the combustion of kerosene goes to heating water.

Efficiency = Q1 * 100% / Q2, where Q1 is the amount of heat that goes to heat the water, Q2 is the amount of water that is released during the combustion of kerosene.

Q1 = C * m1 * (t2 – t1).

Q2 = q * m2.

Efficiency = С * m1 * (t2 – t1) * 100% / q * m2.

m2 = С * m1 * (t2 – t1) * 100% / q * efficiency.

m2 = 4200 J / kg * ° C * 1.2 kg * (100 ° C – 18 ° C) * 100% / 43,000,000 J / kg * 60% = 0.016 kg.

Answer: you need to burn m2 = 0.016 kg.