Question

4. КПД керосиновой горелки 40%. Какую массу керосина надо сжечь на этой, чтобы нагреть 1 кг воды на 75 degrees * C ? Удельная теплоëмкость воды 4200 Дж/(кг. °С), удельная теплота сгорания керосина 45 МДж/кг.​

Answer 1

Ответ: 17,5г

Объяснение:

Дано:

m=1кг

с=4200Дж/кг°С

η=40%=0,4

Δt=75°C

q=45МДж/кг=45000000Дж/кг

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m1-?

Q=cmΔt=4200*1*75=315000Дж

m1=Q/(qη)=315000/(45000000*0,4)=0,0175кг=17,5г

Answer 2

Calculating the Amount of Kerosene Needed to Heat Water

To calculate the amount of kerosene needed to heat 1 kg of water to 75°C with a kerosene burner efficiency of 40%, we can use the following formula:

Q = mcΔT / η

Where: - Q = heat energy required to raise the temperature of the water - m = mass of water - c = specific heat capacity of water - ΔT = change in temperature - η = efficiency of the kerosene burner

Given: - m = 1 kg - c = 4200 J/(kg·°C) - ΔT = 75°C - η = 40% (or 0.40 as a decimal) - Specific heat of combustion of kerosene = 45 MJ/kg (1 MJ = 1,000,000 J)

Let's calculate the amount of kerosene needed using the provided information.

First, we calculate the heat energy required to raise the temperature of the water:

Q = (1 kg) * (4200 J/(kg·°C)) * (75°C) = 315,000 J

Now, we need to account for the efficiency of the kerosene burner. The actual heat energy required will be:

Actual Q = Q / η = 315,000 J / 0.40 = 787,500 J

To find the mass of kerosene needed, we can use the specific heat of combustion of kerosene:

Mass of kerosene = Actual Q / Specific heat of combustion = 787,500 J / (45,000,000 J/kg) = 0.0175 kg

Therefore, to heat 1 kg of water to 75°C with a kerosene burner efficiency of 40%, approximately 0.0175 kg of kerosene is needed.