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Thermodynamics — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics3 Marks
Question
A perfect Carnot engine utilizes an ideal gas. The source temperature is 500K and sink temperature is 375K. If the engine takes 600K cal per cycle from the source, compute:
  1. The efficiency of the engine.
  2. Work done per cycle.
  3. Heat rejected to the sink per cycle.
✓

Answer

Here T1 = 500K

T2 = 375K

Q1 = Heat absorbed per cycle = 600 kcal

  1.  

$\therefore$ Using the relation,

$\eta=1-\frac{\text{T}_2}{\text{T}_1},$ we get

$\eta=\frac{\text{T}_1-\text{T}_2}{\text{T}_1}$

$=\frac{500-375}{500}$

$=\frac{125}{500}=0.25$

$\eta\%=0.25\times100$

$=25\%$

  1. Let W = work done per cycle

$\therefore$ Using the relation

$\eta=\frac{\text{W}}{\text{Q}_1},$ we get

$\text{W}=\eta\text{Q}_1=0.25\times600\text{ kcal}=150\text{ kcal}$

$=150\times10^3\times4.2\text{J}$

$=6.3\times10^5\text{J}$

  1. Let Q2 = Heat rejected to the sink

$\therefore$ Using the relation

$\text{W}=\text{Q}_1-\text{Q}_2,$ we get

$\text{Q}_2=\text{Q}_1-\text{W}$

$=600-150=450\text{ kcal}$

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