Question

Какое количество теплоты потребуется для обращения в пар 200 грамм воды, взятой при температуре 20 С градусов

Answer 1

Q=cm(t2-t1)
Q=4200 Дж\кг * 0.2кг * (100гр-20гр)=67200Дж=67.2кДж

Answer 2

Calculating Heat Required to Convert Water to Steam

To calculate the heat required to convert 200 grams of water at 20°C to steam, we can use the formula:

Q = mcΔT + mL

Where: - Q = heat required - m = mass of the substance (in this case, water) - c = specific heat capacity of the substance - ΔT = change in temperature - L = latent heat of vaporization

Using the given values: - m = 200 grams - c = 4200 J/(kg·°C) (specific heat capacity of water) - ΔT = 80°C (change in temperature from 20°C to 100°C) - L = 2.3 × 10^6 J/kg (latent heat of vaporization for water)

We can calculate the heat required step by step:

1. Q1 = mcΔT - Q1 = 4200 J/(kg·°C) × 0.2 kg × 80°C - Q1 = 67200 J

2. Q2 = mL - Q2 = 2.3 × 10^6 J/kg × 0.2 kg - Q2 = 460000 J

3. Q = Q1 + Q2 - Q = 67200 J + 460000 J - Q = 527200 J

Therefore, the heat required to convert 200 grams of water at 20°C to steam is 527200 Joules.

[[SOURCE #2]]