In each cycle of a carnot engine, 104 j of heat is absorbed from the high-temperature reservoir and 51 j is exhausted to the lowtemperature reservoir. what is the efficiency of the engine?

the efficiency of the Carnot engine is e=0 .5060 (50.96%)

Explanation:

The carnot cycle effifiency e is calculated through

e = W/Q₂

where

W= work

Q₂= heat absorbed

Q₁= heat exhausted

from an energy balance

W= Q₂ - Q₁ ( the proportion of the heat absorbed that was not released was converted to work)

therefore

e=(Q₂-Q₁)/Q₂ = 1 - Q₁/Q₂ = 1 - 51 J/104 J = 0.5060 (50.96%)

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

the answer your looking for is c).