# Efficiency of engine is 40%,at sink temperature is 27 degree celsius.if increase 10% efficiency,how much source temperature ?

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2/5=T₁-300/T₁

T₁=500k

T₁=227degree celsius

if efficiency increased by10 % then

50/100=T₁¹-T₂/T₁¹

1/2=T₁¹-300/T₁¹

T₁¹=600k

=327 degree celsius

T₁¹-T₁=327-227

=100 degree celsius

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Carnot Engine in Thermodynamics:

efficiency η = 40%

T₁ = 27 degC = 300 °K = heat sink temperature

T₂ = ? = temperature of the heat source

40 % =* η = 1 - T₁ / T₂ *= 1 - 300 / T₂

T₂ = 300 / 0.60 = 500 °K =* 227 °C*

If the efficiency is increased by 10% to 50 %

Then 1 - T₁ / T₂ = η = 0.50

T₂ = 2 T₁ = 600 °K =* 327 °C*

*Thus the temperature of the source is to be increased by 100 °C to increase the efficiency by 10 % for an ideal Carnot machine.*

efficiency η = 40%

T₁ = 27 degC = 300 °K = heat sink temperature

T₂ = ? = temperature of the heat source

40 % =

T₂ = 300 / 0.60 = 500 °K =

If the efficiency is increased by 10% to 50 %

Then 1 - T₁ / T₂ = η = 0.50

T₂ = 2 T₁ = 600 °K =