Question

Какое количество теплоты выделится при конденсации 200 г водяного пара, имеющего температуру 100°C, и охлаждении образовавшейся воды до температуры 40°C?

Answer 1

Дано:                             Решение: Q = mL     Q = CmΔt
m = 0.2 кг.                           Q = 0.2 x 2300000 = 460 000 Дж
t1 = 100 °C                         Q = 4200 x 0.2 x 60 = 50 400 Дж
t2 = 40 °C
C = 4200 Дж / кг                        460 + 50.4 = 510.4 кДж
L = 2300000 Дж / кг                     Ответ: 510.4 кДж

Answer 2

Calculation of Heat Released during Condensation and Cooling of Water

To calculate the amount of heat released during the condensation of 200 g of water vapor at 100°C and the subsequent cooling of the formed water to 40°C, we can use the formula:

Q = m * c * ΔT

Where: - Q is the heat released or absorbed (in joules) - m is the mass of the substance (in grams) - c is the specific heat capacity of the substance (in J/g°C) - ΔT is the change in temperature (in °C)

In this case, we need to calculate the heat released during condensation and cooling, so we will consider the heat released during condensation as negative and the heat absorbed during cooling as positive.

1. Heat released during condensation: - Mass of water vapor (m1) = 200 g - Initial temperature (t1) = 100°C - Final temperature (t2) = 100°C (since condensation occurs at constant temperature) - Specific heat capacity of water (c) = 4.18 J/g°C (average value)

Using the formula, we can calculate the heat released during condensation:

Q1 = m1 * c * (t2 - t1)

Substituting the values:

Q1 = 200 g * 4.18 J/g°C * (100°C - 100°C) = 0 J

Therefore, no heat is released during condensation.

2. Heat absorbed during cooling: - Mass of water (m2) = 200 g - Initial temperature (t1) = 100°C - Final temperature (t2) = 40°C - Specific heat capacity of water (c) = 4.18 J/g°C (average value)

Using the formula, we can calculate the heat absorbed during cooling:

Q2 = m2 * c * (t2 - t1)

Substituting the values:

Q2 = 200 g * 4.18 J/g°C * (40°C - 100°C) = -33440 J

Therefore, the heat absorbed during cooling is -33440 J (negative value indicates heat release).

Answer:

The amount of heat released during the condensation of 200 g of water vapor at 100°C and the subsequent cooling of the formed water to 40°C is approximately -33440 J.