Question

Срочно пожалуйста!! Какова масса керосина, которой надо сжечь, чтобы нагреть 2л воды от температуры 20С до кипения? Вода получает 50% теплоты сгорания керосина.Написать с дано.

Answer 1

M1=2 кг
/\t=80 градусов
КПД= 0.5
q =46000000 Дж/кг
0.5=(4200*2*80)/46000000*m
m=0.03 кг

Answer 2

Calculation of the Mass of Kerosene Required to Heat 2 Liters of Water

To calculate the mass of kerosene required to heat 2 liters of water from 20°C to boiling, we need to consider the specific heat capacity of water, the heat of combustion of kerosene, and the fact that water absorbs 50% of the heat released during combustion.

Given: - Volume of water (V) = 2 liters - Initial temperature of water (T1) = 20°C - Final temperature of water (T2) = boiling point (100°C) - Efficiency of heat transfer from kerosene to water = 50%

To calculate the mass of kerosene required, we can use the following steps:

1. Calculate the heat energy required to raise the temperature of the water from T1 to T2 using the formula:

Q = m * c * ΔT

Where: - Q is the heat energy - m is the mass of water - c is the specific heat capacity of water - ΔT is the change in temperature

The specific heat capacity of water is approximately 4.18 J/g°C.

2. Calculate the heat energy released during the combustion of kerosene using the formula:

Q = m * H

Where: - Q is the heat energy - m is the mass of kerosene - H is the heat of combustion of kerosene

The heat of combustion of kerosene is approximately 46 * 10^6 J/kg.

3. Since the water absorbs only 50% of the heat released during combustion, we need to multiply the heat energy released by 2.

4. Equate the two heat energy equations and solve for the mass of kerosene (m).

Now let's perform the calculations:

1. Calculate the heat energy required to raise the temperature of the water:

Q1 = m_water * c_water * ΔT

- m_water = V_water * ρ_water - V_water = 2 liters = 2000 cm^3 - ρ_water = density of water = 1 g/cm^3 - c_water = specific heat capacity of water = 4.18 J/g°C - ΔT = T2 - T1 = 100°C - 20°C = 80°C

Substituting the values:

Q1 = (2000 cm^3) * (1 g/cm^3) * (4.18 J/g°C) * (80°C)

2. Calculate the heat energy released during the combustion of kerosene:

Q2 = m_kerosene * H_kerosene

- H_kerosene = heat of combustion of kerosene = 46 * 10^6 J/kg

Substituting the values:

Q2 = m_kerosene * (46 * 10^6 J/kg)

3. Multiply the heat energy released by 2 to account for the 50% absorption by water:

Q2 = 2 * Q2

4. Equate Q1 and Q2:

(2000 cm^3) * (1 g/cm^3) * (4.18 J/g°C) * (80°C) = 2 * m_kerosene * (46 * 10^6 J/kg)

Now we can solve for the mass of kerosene (m_kerosene):

m_kerosene = [(2000 cm^3) * (1 g/cm^3) * (4.18 J/g°C) * (80°C)] / [2 * (46 * 10^6 J/kg)]

Using the given values and performing the calculations:

m_kerosene ≈ 0.16 kg

Therefore, approximately 0.16 kg of kerosene is required to heat 2 liters of water from 20°C to boiling, considering that the water absorbs 50% of the heat released during combustion.

Please note that the above calculations are based on the given information and assumptions about the specific heat capacity of water, the heat of combustion of kerosene, and the efficiency of heat transfer.