Question

В идеальной тепловой машине за счет за каждого килоджоуля энергии,получаем о и от нагревателя,совершается работа 400 дж.Определите КПД маины и температууру холодильника,если температура нагревателя 500 К

Answer 1

Qn=1 кДж =1000 Дж

A =400 Дж

кпд = A / Qn = 400/1000 =0.4 =40 %

кпд =1 - Тх/Tn

Tx = (1 - кпд) *Tn = (1-0.4)*500 =300 K

Answer 2

Calculation of Efficiency (КПД) and Temperature of the Refrigerator

To determine the efficiency (КПД) of the ideal heat engine and the temperature of the refrigerator, we can use the given information:

- The work done by the heat engine is 400 J. - The energy obtained from the heater is 1 kilojoule (1000 J).

The efficiency (КПД) of a heat engine is defined as the ratio of the work done by the engine to the energy obtained from the heater. Mathematically, it can be expressed as:

КПД = (Work done by the engine / Energy obtained from the heater) * 100%

Substituting the given values:

КПД = (400 J / 1000 J) * 100%

Calculating the efficiency:

КПД = 40%

Therefore, the efficiency of the ideal heat engine is 40%.

To determine the temperature of the refrigerator, we need additional information. The temperature of the heater is given as 500 K, but the temperature of the refrigerator is not provided. Without this information, we cannot calculate the temperature of the refrigerator.

Please provide the temperature of the refrigerator so that we can calculate it accurately.