Answer: 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
