Question

a heat engine operates between 2100k and 700k it's actual efficiency is 40% what percentage of its maximum possible efficiency is this

Answer 1

Answer:

CARNOT ENGINE

Explanation:

♣ THERMODYNAMICS ♥

★ GIVEN ;

» Temperature of Source ( T¹ ) = 2100 K

» Temperature of Sink ( T² ) = 700 K

★ ORIGINAL EFFICIENCY ≈ 40 % ★

→★ MAXIMUM EFFICIENCY ( ẞ ) ≈ ???

★ BY USING FORMULA ;

$\beta \: = \: ( \: 1 - \frac{ {t}^{2} }{ {t}^{1} } )$

★ ẞ = 1 - ( 700 / 2100 )

» ẞ = ( 2100 - 700 ) / 2100

» ẞ = 1400 / 2100

★» ẞ = 2 / 3

★ %ẞ = 2 / 3 × 100 = 66% ★

_________________________________________

ANSWER :- Hence ; Maximum % EFFICIENCY of CARNOT ENGINE is ;

★ ẞ = 66 % ★

Answer 2

Answer:60

Explanation:

The condition for efficiency in case of Carnot Engine is e = 1- (TL/TH) = 1-(700/2100) = 66%=Theoretical Efficiency

But actual efficiency is 40% which is 60% of the theoretical efficiency 66*60/100= 39.6%

So the answer is 60%