Question

Сколько теплоты выделится при конденсации 300г спирта , взятого при температуре кипения , и дальнейшего охлаждения его до 18градусов

Answer 1

Т.к. по условию процесс происходит при t кип, можно сразу переходить к конденсации Q1=Lm, где L-удельная теплота парообразования, равная 0.9*10^6 Дж/кг, m-масса, равная 0.3 кг
Q1=0.27*10^6Дж
Далее - процесс охлаждения от t кип до 18 градусов. t кип спирта = 78 градусов
Q2=cm(tкип-t)= 2500 Дж/(кг*градус C)*0.3кг*60 град С= 45*10^3Дж
Qобщ=Q1+Q2=315 кДж
на всякий случай сверьте справочные данные и расчеты 

Answer 2

Calculating Heat Released During Condensation of 300g of Alcohol

To calculate the heat released during the condensation of 300g of alcohol at its boiling temperature and subsequent cooling to 18 degrees Celsius, we can use the formula:

Q = Lm

Where: - Q = Heat released - L = Latent heat of condensation - m = Mass of the substance

Using the provided information, we can calculate the heat released step by step.

1. Calculating the Heat Released During Condensation: - Given: L = 2.3*10^6 J/kg, m = 300g = 0.3kg - Using the formula: Q = Lm - Substituting the values: Q = 2.3*10^6 * 0.3 - Calculating: Q = 6.9*10^5 J

The heat released during the condensation of 300g of alcohol is 6.9*10^5 Joules.

2. Calculating the Heat Released During Cooling to 18°C: - The specific heat capacity of alcohol is not provided in the search results. Therefore, the exact heat released during the cooling process cannot be calculated with the given information.

In conclusion, the heat released during the condensation of 300g of alcohol at its boiling temperature is 6.9*10^5 Joules. However, the exact heat released during the subsequent cooling to 18 degrees Celsius cannot be determined without the specific heat capacity of alcohol.