A Carnot’s engine has 25% efficiency with a sink at 9^oC. By how many degrees should the temperature of the source be increased in order to raise the efficiency to 50%?

Solution

Given,
1^st Case:
Efficiency ([latex]\eta[/latex]) = 25% = 0.25
Temperature of sink (T₂) = 9^oC = 9 + 273 = 282 K
We know that,
Or, 0.25 = 1 – [latex]\frac{282}{T_1}[/latex]
Or, T₁ = Temperature of source = 376K
2^nd Case:
Let, temperature of source is to be increased by x K.
So, [latex]\eta = 1 – \frac{T_2}{T_1}[/latex]

Or, 0.5 = 1 – [latex]\frac{282}{376+x}[/latex]

            Or, x = 188 K

For a carnot ideal engine, temperature of sink at temperature 27^oC and source is at temperature 127^oC. Calculate its efficiency.

Solution

Temperature of sink ([latex]T_2[/latex]) = 27^oC = 27 + 273 = 300 K
Temperature of source (T₁) = 127^oC = 127 + 273 = 400 K
Efficiency ([latex]\eta[/latex]) =?
We know that,
[latex]\eta = (1-\frac{T_2}{T_1})\times 100%[/latex]

= [latex](1-\frac{300}{400})\times 100%[/latex]

                        = 25%

A refrigerator has a coefficient of performance of 1.95. In each cycle, it absorbs [latex]3\times 10^4[/latex] J of heat from cold reservoir. How much heat is discarded to the high temperature during each cycle?

Solution

Coefficient of performance ([latex]\beta[/latex]) = 1.95
For refrigerator,
Heat absorbed (Q₂) = [latex]3\times 10^4 J[/latex]
Heat discarded (Q₁) =?
We know that,
[latex] \beta = \frac{\text{heat absorbed } (Q_2)}{\text{work done }(Q_1 – Q_2)}[/latex]

Or, 1.95 = [latex]\frac{3\times 10^4}{Q_1 – Q_2}[/latex]

Or, [latex]Q_1 – Q_2 = \frac{3\times 10^4}{1.95} [/latex]

= [latex]1.54\times 10^4[/latex]

Or, [latex]Q_1 = (1.54\times 10^4) + Q_2[/latex]

[latex]= (1.54\times 10^4) + (3\times 10^4)[/latex]

Or, [latex]Q_1 = 4.54\times 10^4 J[/latex]

Two carnot engines A & B have their sources at 400 K and 350 K, and sinks at 350 K and 300 K respectively. Which engine is more efficient and by how much?

Solution