Question

Помогите пожалуйста!очень прошу...Дано: А=3,5Кдж Q2=3 кдж, Tхолодильника=27С . НАЙТИ:А)КПД ? Б)Tнагревателя ? отдааю 25 баллов!

Answer 1

Прошу прощения за почерк.

Answer 2

Given Information:

- A = 3.5 kJ (amount of energy absorbed by the refrigerator) - Q2 = 3 kJ (amount of energy released by the refrigerator) - Tхолодильника = 27°C (temperature of the refrigerator)

A) Finding the Efficiency (КПД):

The efficiency (КПД) of a refrigerator can be calculated using the formula:

КПД = (Q2 / A) * 100%

Substituting the given values:

КПД = (3 kJ / 3.5 kJ) * 100%

Calculating the efficiency:

КПД = 0.857 * 100%

Therefore, the efficiency of the refrigerator is approximately 85.7%.

B) Finding the Temperature of the Heater (Tнагревателя):

To find the temperature of the heater (Tнагревателя), we need to use the formula for the efficiency of a refrigerator:

КПД = (Tнагревателя - Tхолодильника) / Tнагревателя

Substituting the given values and the efficiency (КПД) calculated in part A:

0.857 = (Tнагревателя - 27°C) / Tнагревателя

Simplifying the equation:

0.857 * Tнагревателя = Tнагревателя - 27°C

0.857 * Tнагревателя - Tнагревателя = -27°C

-0.143 * Tнагревателя = -27°C

Solving for Tнагревателя:

Tнагревателя = (-27°C) / (-0.143)

Calculating the temperature of the heater:

Tнагревателя ≈ 188.811°C

Therefore, the temperature of the heater is approximately 188.811°C.

I hope this helps! Let me know if you have any further questions.