suatu mesin carnot, jika reservoir panasnya bersuhu 400k akan mempunyai efesiensi 40% jika reservoir panasnya bersuhu 640k berapa % efisiensinya

Efisiensi = η

η =1 -(T2/T1)
40% = 1- (T2/400)
0,4 = 1- (T2/400)
(T2/400)= 0,6
T2 = 0,6*400 = 240 K

η =1 -(T2/T1)
η = 1- (240/640)
η= 1- (⅜)
η =⅝
η =⅝*100% = 62,5 %

TERMODINAMIKA
• Mesin Carnot

Keadaan pertama
η = 40 % = 0,4
T₁ = 400 K
T₂ = __?

η = 1 - T₂/T₁
0,4 = 1 - T₂/400
T₂/400 = 1 - 0,4
T₂/400 = 0,6
T₂ = 0,6 • 400
T₂ = 240 K


Keadaan kedua
T₁' = 640 K ← diubah
T₂ = 240 K ← tetap
η' = ___?

η' = 1 - T₂/T₁'
η' = 1 - 240/640
η' = 1 - ⅜
η' = = 5/8 = 62,5 % ✔️