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In order for heat to transfer in a reversible process, the temperature difference must be approximately zero. Therefore at the isotherms in a carnot engine, there is negligible temperature difference. How does the temperature of the (surroundings) from source to sink then suddenly change after a sudden adiabatic expansion/compression?

If the temperature of the source and sink are constant, then the efficiency of the ideal carnot engine comes to 0. Then how come this is better than the otto cycle? Additionally, I'm slightly confused as to what the temperature of source and sink is in an otto cycle, since the temperature of the surroundings continuously change during the isochoric parts.

To my understanding, the carnot cycle consists of isothermal expansion, where work is done by adding heat to the system, this is done at negligible temperature difference ideally. (rapid) Adiabatic expansion, reducing temperature. Isothermal compression, where heat is taken away and negative work is done, followed by adiabatic compression. It is possible to derive efficiency in terms of temperature by mathematically computing the expression $$\frac{W}{Q_h}$$ to get $$1-\frac{T_c}{T_h}$$. But how can $$T_c$$ be different from $$T_h$$?

This could be solved by rapidly transferring the engine into another reservoir?

In order for heat to transfer in a reversible process, the temperature difference must be approximately zero. Therefore at the isotherms in a carnot engine, there is negligible temperature difference. How does the temperature of the (surroundings) from source to sink then suddenly change after a sudden adiabatic expansion/compression?

If the temperature of the source and sink are constant, then the efficiency of the ideal carnot engine comes to 0. Then how come this is better than the otto cycle? Additionally, I'm slightly confused as to what the temperature of source and sink is in an otto cycle, since the temperature of the surroundings continuously change during the isochoric parts.

In order for heat to transfer in a reversible process, the temperature difference must be approximately zero. Therefore at the isotherms in a carnot engine, there is negligible temperature difference. How does the temperature of the (surroundings) from source to sink then suddenly change after a sudden adiabatic expansion/compression?

If the temperature of the source and sink are constant, then the efficiency of the ideal carnot engine comes to 0. Then how come this is better than the otto cycle? Additionally, I'm slightly confused as to what the temperature of source and sink is in an otto cycle, since the temperature of the surroundings continuously change during the isochoric parts.

To my understanding, the carnot cycle consists of isothermal expansion, where work is done by adding heat to the system, this is done at negligible temperature difference ideally. (rapid) Adiabatic expansion, reducing temperature. Isothermal compression, where heat is taken away and negative work is done, followed by adiabatic compression. It is possible to derive efficiency in terms of temperature by mathematically computing the expression $$\frac{W}{Q_h}$$ to get $$1-\frac{T_c}{T_h}$$. But how can $$T_c$$ be different from $$T_h$$?

This could be solved by rapidly transferring the engine into another reservoir?

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# Efficiency of reversible carnot engine must be 0?

In order for heat to transfer in a reversible process, the temperature difference must be approximately zero. Therefore at the isotherms in a carnot engine, there is negligible temperature difference. How does the temperature of the (surroundings) from source to sink then suddenly change after a sudden adiabatic expansion/compression?

If the temperature of the source and sink are constant, then the efficiency of the ideal carnot engine comes to 0. Then how come this is better than the otto cycle? Additionally, I'm slightly confused as to what the temperature of source and sink is in an otto cycle, since the temperature of the surroundings continuously change during the isochoric parts.