30
Problem solving
Textbook: p. 86
GoalSolve problems on the efficiency, work and heat of heat engines.
New words
useful work · foydali ishheat taken from the heater · isitkichdan olingan issiqlikheat given to the cooler · sovitkichga berilgan issiqlikmaximum efficiency · eng katta FIK
Explanation
In engine problems tell apart four quantities: Q₁ (taken from the heater), Q₂ (given to the cooler), A = Q₁ – Q₂ and η = A/Q₁. If two are known you can find the others: Q₁ = A/η, Q₂ = Q₁ – A = A(1 – η)/η. If the engine is ideal (Carnot), η = 1 – T₂/T₁ and temperatures must be in kelvins. If η is given in percent, divide by 100 in the calculation (30 % = 0.3). Check the result: η < 1, Q₁ > A, Q₂ > 0. When comparing two engines, a gain in efficiency comes from cooling the cooler or heating the heater.
Worked examples
In one cycle an engine does 300 J of work and gives 900 J to the cooler: Q₁ = A + Q₂ = 1200 J; η = 300/1200 = 0.25 = 25 %.
An ideal engine: T₁ = 600 K, T₂ = 300 K, Q₁ = 1000 J per cycle. η = 1 – 300/600 = 0.5; A = ηQ₁ = 500 J; Q₂ = Q₁ – A = 500 J.
Class activity
“Find the missing parts” game: the teacher gives two of Q₁, Q₂, A, η; the next group finds the other two and, if correct, poses the next problem.
Practice
1
An engine of efficiency 20 % does 4 kJ of work per cycle. How many kJ does it take from the heater (Q₁)?
20
2
How many kJ does the same engine give to the cooler per cycle?
16
3
A Carnot engine has efficiency 75 % and the cooler is at 250 K. What is the heater temperature in K?
1000
4
T₁ = 500 K, T₂ = 300 K. Which raises the efficiency more: raising T₁ by 100 K or lowering T₂ by 100 K?
Lowering T₂: η = 1 – 200/500 = 60 %; raising T₁: 1 – 300/600 = 50 %.