Question

Какое количество теплоты выделяется при превращении 400г пара имеющего температуру 100^ в лёд при -20

Answer 1

Формула Q=LM. Где Q - количество теплоты, выделившейся при конденсации. L - удельная теплота парообразования/конденсации. М - масса пара. L=2 300 000 Дж/Кг. получается, что Q= 2 300 000 Дж/Кг*0,4Кг = 920 Кдж.

Answer 2

Calculation of Heat Released during the Transformation of 400g of Steam at 100°C to Ice at -20°C

To calculate the amount of heat released during the transformation of 400g of steam at 100°C to ice at -20°C, we can use the formula:

Q = mcΔT + mL

Where: - Q is the heat released or absorbed during the transformation - m is the mass of the substance (in this case, steam) - c is the specific heat capacity of the substance - ΔT is the change in temperature - L is the latent heat of transformation

Let's break down the calculation step by step:

1. Calculate the heat released when the steam cools down from 100°C to 0°C: - The specific heat capacity of water is approximately 4200 J/(kg·°C). - The mass of the steam is 400g, which is equal to 0.4kg. - The change in temperature is 100°C - 0°C = 100°C. - Using the formula Q = mcΔT, we can calculate the heat released during this step: Q1 = (0.4kg) * (4200 J/(kg·°C)) * (100°C) = 16,800 J.

2. Calculate the heat released when the steam condenses into water at 0°C: - The latent heat of vaporization (or condensation) for water is approximately 2.3 * 10^6 J/kg. - The mass of the steam is 400g, which is equal to 0.4kg. - Using the formula Q = mL, we can calculate the heat released during this step: Q2 = (0.4kg) * (2.3 * 10^6 J/kg) = 920,000 J.

3. Calculate the heat released when the water cools down from 0°C to -20°C: - The specific heat capacity of ice is approximately 2100 J/(kg·°C). - The mass of the water is 400g, which is equal to 0.4kg. - The change in temperature is 0°C - (-20°C) = 20°C. - Using the formula Q = mcΔT, we can calculate the heat released during this step: Q3 = (0.4kg) * (2100 J/(kg·°C)) * (20°C) = 16,800 J.

4. Sum up the heat released during each step to get the total heat released: Q = Q1 + Q2 + Q3 = 16,800 J + 920,000 J + 16,800 J = 953,600 J.

Therefore, the total amount of heat released during the transformation of 400g of steam at 100°C to ice at -20°C is approximately 953,600 J.

Please note that this calculation assumes no heat loss to the surroundings and that the specific heat capacities and latent heat values used are approximate averages.