In the carnot engine, when heat is absorbed from the source, its temperature

In ta carnot engine, when heat is absorbed from the source, its temperature

The efficiency of a carnot heat engine is 0.6 when the absolute temperatures of the source and sink are T_(1) and T_(2) respectively. The efficiency of another carnot engine is also 0.6 but the source and sink temperatures are different. What are the temperatures of the source and the sink of the other engine ?

Carnot engine is an ideal heat engine, which converts heat energy into mechanical energy. Efficiency of Carnot engine is given by eta =1 - (T_2 // T_1) , where T_1 is temperature of source and T_1 , is temperature of sink. If Q_1 is the amount of heat absorbed/cycle from the source, Q_2 is the amount of heat rejected/cycle to the sink and W is the amount of useful work done/cycle, then W= Q_1 -Q_2 and (Q_2)/(Q_1) = (T_2)/(T_1) Work done/cycle by the engine in the above question is. if source and sink temperature respectively 227 and 127 . heat absorbed by engine is given 6 .

Carnot engine is an ideal heat engine, which converts heat energy into mechanical energy. Efficiency of Carnot engine is given by eta =1 - (T_2 // T_1) , where T_1 is temperature of source and T_1 , is temperature of sink. If Q_1 is the amount of heat absorbed/cycle from the source, Q_2 is the amount of heat rejected/cycle to the sink and W is the amount of useful work done/cycle, then W= Q_1 -Q_2 and (Q_2)/(Q_1) = (T_2)/(T_1) A Carnot engine absorbs 6 xx 10^5 " cal. at "227^@ C. Heat rejected to the sink at 127^@ C is

A perfect Carnot engine utilises an ideal gas as the working substance. The source temperature is 227^circC and the sink temperature is 127^circC . Find the efficiency of the engine, and find the heat received from the source and the heat released to the sink when 10,000 J of external work is done.

Carnot engine is an ideal heat engine, which converts heat energy into mechanical energy. Efficiency of Carnot engine is given by eta =1 - (T_2 // T_1) , where T_1 is temperature of source and T_1 , is temperature of sink. If Q_1 is the amount of heat absorbed/cycle from the source, Q_2 is the amount of heat rejected/cycle to the sink and W is the amount of useful work done/cycle, then W= Q_1 -Q_2 and (Q_2)/(Q_1) = (T_2)/(T_1) Efficiency of the engine in the above question is .if source and sink temperature respectively 227 and 127 .

A Carnot engine, having an efficiency of eta= 1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is