A Carnot engine has an efficiency of 0.610, and the temperature of its cold reservoir is376 K.(a) Determine the temperature of its hotreservoir.(b) If 5320 J of heat is rejected to thecold reservoir, what amount of heat is put into the engine?

a) Th = 935.89 K, b) Qh = 13,241.8 J

Explanation: efficiency of a carnot engine is given below as

E = 1 - (Tc/Th)

Where Tc = temperature at the cold reservoir and Th = temperature the hot reservoir.

From the question, E = 0.610, Tc = 376k, Th =?

By substituting parameters, we have

0.610 = 1 - (376/Th)

(376/Th) = 1 - 0.610

376/Th = 0.39

376 = 0.39 × Th

Th = 365/0.39

Th = 935.89 K

Question B)

Heat transfer in a carnot engine is defined mathematically below as

|Qc|/|Qh| = Tc/Th

The modulus sign beside Qc and Qh denotes that their values can not be negative

Qc = heat rejected at cold reservoir = 5320 J

Qh = heat put into the engine =?

Tc = temperature at cold reservoir = 376 K

Th = temperature at hot reservoir = 935.89 K.

By substituting the parameters, we have that

5320/Qh = 376/935.89

By cross multiplication

Qh×376 = 5320×935.89

Qh = 5320×935.89/376

Qh = 4,978,934.8/376

Qh = 13,241.8 J

vapour or gas assuming atmospheriv pressure

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)