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

see earlier answer

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.