Question

В бак с водой при 30С впустили 400г стоградусного пара.После конденсации пара температура установилась 32С.Какое количество воды было в баке?

Answer 1

М=(Lm-cm(100-32))/(32-30)=(0,4L-4200*0,4*68)/2, удельная теплоту парообразования глянь в таблице,нет под рукой.

Answer 2

Calculation of the Amount of Water in the Tank

To determine the amount of water in the tank, we need to consider the change in temperature and the condensation of the steam.

Given: - Initial temperature of the water in the tank: 30°C - Mass of the steam introduced into the tank: 400g - Final temperature after condensation: 32°C

To solve this problem, we can use the principle of conservation of energy. The heat gained by the water is equal to the heat lost by the steam during condensation.

The heat gained by the water can be calculated using the formula:

Q1 = m1 * c1 * ΔT1

Where: - Q1 is the heat gained by the water - m1 is the mass of the water - c1 is the specific heat capacity of water (approximately 4.18 J/g°C) - ΔT1 is the change in temperature of the water (32°C - 30°C = 2°C)

The heat lost by the steam during condensation can be calculated using the formula:

Q2 = m2 * L

Where: - Q2 is the heat lost by the steam - m2 is the mass of the steam - L is the latent heat of vaporization of water (approximately 2260 J/g)

Since the heat gained by the water is equal to the heat lost by the steam, we can set up the equation:

Q1 = Q2

m1 * c1 * ΔT1 = m2 * L

We know the mass of the steam (m2) is 400g. We need to find the mass of the water (m1).

Let's solve the equation for m1:

m1 = (m2 * L) / (c1 * ΔT1)

Substituting the known values:

m1 = (400g * 2260 J/g) / (4.18 J/g°C * 2°C)

Calculating this expression gives us:

m1 ≈ 5423.44g

Therefore, the amount of water in the tank is approximately 5423.44 grams.

Please note that this calculation assumes that there is no heat exchange with the surroundings and that the specific heat capacity and latent heat of vaporization remain constant over the given temperature range.