Question

Сколько бензина нужно сжечь, чтобы испарить 300 грамм воды при температуре 100 градусов. РЕШИТЕ СРОЧНО, НУЖНО С ДАНО И ФОРМУЛАМИ​

Answer 1

Ответ:

Надо сжечь 8.076 г. бензина

Q1+Q2=Q

Answer 2

#### Calculation of the Amount of Gas Required to Evaporate 300 Grams of Water at 100 Degrees Celsius To calculate the amount of gas required to evaporate 300 grams of water at 100 degrees Celsius, we need to consider the heat required to raise the temperature of the water to its boiling point and then the heat required for the phase change from liquid to gas. 1. Heat Required to Raise the Temperature of Water: - The specific heat capacity of water is approximately 4.18 J/g°C. - The temperature change is from 100°C (initial temperature) to 100°C (boiling point). - The formula to calculate the heat required is: **Q = m * c * ΔT**, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the temperature change. - Substituting the values, we get: **Q = 300 g * 4.18 J/g°C * (100°C - 100°C) = 0 J**. 2. Heat Required for Phase Change: - The heat required for the phase change from liquid to gas is known as the heat of vaporization. - The heat of vaporization for water is approximately 40.7 kJ/mol or 2260 J/g. - The formula to calculate the heat required is: **Q = m * H**, where Q is the heat, m is the mass, and H is the heat of vaporization. - Substituting the values, we get: **Q = 300 g * 2260 J/g = 678,000 J**. 3. Total Heat Required: - To find the total heat required, we add the heat required to raise the temperature of the water and the heat required for the phase change. - Total heat required = Heat required to raise the temperature + Heat required for phase change = 0 J + 678,000 J = 678,000 J. 4. Conversion of Energy to Gasoline: - Gasoline has an energy density of approximately 31.5 MJ/L or 31,500,000 J/L. - To calculate the volume of gasoline required, we divide the total heat required by the energy density of gasoline. - Volume of gasoline required = Total heat required / Energy density of gasoline = 678,000 J / 31,500,000 J/L = 0.0215 L. Therefore, approximately 0.0215 liters of gasoline would be required to evaporate 300 grams of water at 100 degrees Celsius. Please note that this calculation assumes ideal conditions and does not take into account factors such as heat loss or efficiency.