Question

ПОМОГИТЕ ПОЖАЛУЙСТА!!! СРОЧНО! Задача по физике. Используя КПД спиртовки 37%, рассчитайте объем спирта, который необходимо сжечь, чтобы нагреть на спиртовке 5 л воды от 20°С до 70°С.

Answer 1

Рассматриваю этиловый спирт, у которого q = 3*10^7 Дж/кг
плотность беру 790 кг/м^3

n = Qпол / Qзат = с p V (70 - 20) / m q

n m q = 50 c p V

m = 50 c p V / n q = 5*10 * 42*10^2 * 10^3 * 5*10^-3 / 37*10^-2 * 3*10^7 = 105*10^3 / 111*10^5 = 0,00945 кг

значит, V = m / p = 0,00945 / 790 = 1,196*10^-5 м^3

Answer 2

Calculation of the Amount of Alcohol Needed to Heat 5 Liters of Water

To calculate the volume of alcohol needed to heat 5 liters of water from 20°C to 70°C using a spirit burner with an efficiency of 37%, we can use the following steps:

Step 1: Calculate the energy required to heat the water. The energy required to heat a substance can be calculated using the formula:

Q = m * c * ΔT

Where: - Q is the energy required (in joules) - m is the mass of the substance (in kilograms) - c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius) - ΔT is the change in temperature (in degrees Celsius)

In this case, we have 5 liters of water, which is equivalent to 5000 grams (since 1 liter of water is equal to 1000 grams). The specific heat capacity of water is approximately 4.18 joules per gram per degree Celsius.

Using the formula, we can calculate the energy required to heat the water from 20°C to 70°C:

Q = 5000g * 4.18 J/g°C * (70°C - 20°C)

Step 2: Calculate the energy output of the spirit burner. The energy output of the spirit burner can be calculated by multiplying the energy input (the amount of alcohol burned) by its efficiency. The efficiency is given as 37%.

Let's assume the volume of alcohol burned is V liters.

The energy output can be calculated using the formula:

Energy Output = Energy Input * Efficiency

Step 3: Equate the energy input and output. Since energy input equals energy output, we can set up the following equation:

Energy Input = Energy Output

Substituting the values from Step 1 and Step 2 into the equation, we get:

V * 37% = 5000g * 4.18 J/g°C * (70°C - 20°C)

Simplifying the equation, we have:

V * 0.37 = 5000g * 4.18 J/g°C * 50°C

Now we can solve for V, the volume of alcohol needed.

Let's calculate the result.

Using the given values: - Specific heat capacity of water (c) = 4.18 J/g°C - Efficiency of the spirit burner = 37% - Change in temperature (ΔT) = 70°C - 20°C = 50°C - Mass of water (m) = 5000g

Substituting these values into the equation, we have:

V * 0.37 = 5000g * 4.18 J/g°C * 50°C

Simplifying the equation:

V * 0.37 = 1045000 J

Dividing both sides of the equation by 0.37:

V = 1045000 J / 0.37

Calculating the result:

V ≈ 2,820,270.27 liters

Therefore, approximately 2,820,270.27 liters of alcohol would be needed to heat 5 liters of water from 20°C to 70°C using a spirit burner with an efficiency of 37%.

Please note that this calculation assumes ideal conditions and does not take into account factors such as heat losses or variations in the efficiency of the spirit burner.