A Carnot engine operates between a hot reservoir at 320 degrees Kelvin and a cold reservoir at 260 degrees Kelvin. If it absorbs 500 Joules of heat at the hot reservoir, how much work does it deliver

Leonard Carnot proposed the Carnot engine, a hypothetical thermodynamic cycle. It calculates the highest feasible efficiency that a heat engine can have when converting heat into work and when operating between two reservoirs.
Any system working between [tex]T_1[/tex] (hot reservoir) and [tex]T_2[/tex] (cold reservoir) will never have better efficiency than the Carnot engine running between the identical reservoirs.
Additionally, the efficiency of this form of engine is only reliant on the temperature of the hot and cold reservoirs and is independent of the composition of the working substance.

Solution:

[tex]T_c = 260 K\\ T_h = 320K\\ Q_h = 500J[/tex]

As we know that Carnot engine is ideal, the efficiency depends only on the temperature of the reservoirs(in kelvin scale):

[tex]n = 1 - \frac{T_c}{T_h} \\n = 1 - \frac{260}{320}[/tex]

[tex]n = 1- 0.8125\\n = 0.1875[/tex]

Also, we know that the work done is equal to product of efficiency and heat at hot reservoir:

[tex]W = n\times Q_h\\W = 0.1875 \times 500\\W = 93.75 J[/tex]

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