Question

На сколько изменится энтропия системы при превращении в пар 0,05 кг кипящей при 100оС воды? Удельная теплота парообразования воды 2,26 МДж/кг.

Answer 1

DS=L*m/T=2,26*10^6/373=6*10^3 Дж/К

Answer 2

Calculating Change in Entropy

To calculate the change in entropy of the system when 0.05 kg of boiling water at 100°C is transformed into steam, we can use the formula:

ΔS = q / T

Where: - ΔS is the change in entropy - q is the heat added or removed - T is the temperature in Kelvin

The specific heat of vaporization of water is given as 2.26 MJ/kg.

Calculation

Given: - Mass of water, m = 0.05 kg - Specific heat of vaporization of water, q = 2.26 MJ/kg - Temperature, T = 100°C

First, we need to convert the temperature to Kelvin: T(K) = T(°C) + 273.15

Plugging in the values: T(K) = 100 + 273.15 = 373.15 K

Now, we can calculate the change in entropy using the formula: ΔS = q / T

ΔS = 2.26 MJ / 0.05 kg / 373.15 K

Result

The change in entropy of the system when 0.05 kg of boiling water at 100°C is transformed into steam is calculated to be 0.0121 MJ/K.