A Carnot engine is held between a furnace with a constant temperature of 400° C and the outside environment at a constant temperature of 30° C. There is a leak in the system which allows heat to escape directly from the furnace to the environment. The efficiency of the engine, including the effect of the leak, is 0.25. What is the ratio of the heat that flows into the Carnot engine (QH1) to the heat that flows directly to the environment (QH2)?

Respuesta :

Answer:

[tex]\frac{4}{3}[/tex]

Explanation:

Heat engine converts heat energy into work. Heat flows into the Carnot engine from the furnace and some of the heat is converted into work while the rest is expelled to the environment.

The efficiency of the Carnot engine = Work done by the engine ÷ Heat from the hot reservoir

                                       η [tex]= \frac{W}{Q_{H}}[/tex]

Carnot engine is the most efficient heat engine because it consists of isothermal and adiabatic processes (reversible processes). However, it is not capable of converting all of the input heat into work. The leak decreases its efficiency.

                      Carnot Efficiency [tex]= \frac{T_{H} - T_{C}}{T_{H}}[/tex]

                      Carnot Efficiency [tex]= 1 -\frac{T_{C}}{T_{H}}[/tex]

Where;                                            

[tex]T_{H}[/tex] = The temperature of the hot reservoir

[tex]T_{C}[/tex] = The temperature of the cold reservoir  

                        0.25 [tex]= 1 -\frac{T_{C}}{T_{H}}[/tex]

                        [tex]\frac{T_{C}}{T_{H}}[/tex] = 1 - 0.25

                        [tex]\frac{T_{C}}{T_{H}}[/tex] = 0.75

                        [tex]\frac{T_{C}}{T_{H}}[/tex] =   [tex]\frac{75}{100}[/tex]

                        [tex]\frac{T_{C}}{T_{H}}[/tex] = [tex]\frac{3}{4}[/tex]

                        [tex]\frac{T_{H}}{T_{C}}[/tex] = [tex]\frac{4}{3}[/tex]

Since a Carnot engine is a perfect heat engine,

The temperature of the hot reservoir, [tex]T_{H}[/tex] = The heat that flows into the Carnot engine [tex]Q_{H1}[/tex]

The temperature of the cold reservoir

, [tex]T_{C}[/tex] = The heat that flows directly to the environment [tex]Q_{H2}[/tex]

                        [tex]\frac{Q_{H1}}{Q_{H2}}=[/tex]  [tex]\frac{4}{3}[/tex]

             

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