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Question : 101 of 160
Marks:
+1 ,
-0
Solution:
The given situation is shown below
T 2 ‌ ‌ = 200 K T 3 ‌ ‌ = 350 K T 4 ‌ ‌ = 250 K where,
T 1 , T 2 , T 3 and
T 4 are the temperature of given parts in figure
As we know that,
η c = ‌ = ‌ where,
Q 1 , Q 2 are the heat of part 1,2 respectively and
η C is efficiency of Carnot engine.
∴ ‌ ‌ = ‌ = ‌ ‌ Q 1 − Q 2 = ‌ ⇒ ‌ Q 2 = Q 1 ∕ 2 ‌ and ‌ ‌ η e = ‌ = ‌ = ‌ ⇒ ‌ W = Q 1 ∕ 2 Coeffficient of performance of refrigerator is given as
‌ COP ‌ = ‌ = ‌ = ‌ . . . (i)
where,
T C and
T H are cold and hot temperature i.e. similar to
T 4 and
T 3 , respectively.
∴ ‌ ‌ C O P = ‌ = ‌ = ‌ ⇒ ‌ ‌ ‌ = ‌ ⇒ ‌ ‌ ‌ = ‌ . . . (ii)
Substituting
W = Q 1 ∕ 2 in Eq. (ii), we get
‌ = ‌ ⇒ ‌ ‌ Q 4 ∕ Q 1 = 5 ∕ 4 ⇒ ‌ ‌ Q 4 = ( 5 ∕ 4 ) Q 4 Substituting in Eq. (i), we get
‌ ‌ ‌ = ‌ = ‌ ⇒ ‌ ‌ ‌ ‌ ‌ = ‌ ⇒ ‌ ‌ 5 Q 3 − ‌ Q 1 ‌ ‌ = ‌ ⇒ ‌ ‌ ‌ ‌ ‌ = ‌ = 1.75 ⇒ ‌ ‌ Q 3 ‌ ‌ = ‌ ⇒ ‌ ‌ Q 2 © examsiri.com
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