Respuesta :
5020 Joules.
The maximum theoretical efficiency of an heat engine is
e = 1 - Tc/Th
where
e = efficiency
Tc = cold temperature (absolute)
Th = hot temperature (absolute)
So let's convert the temperatures from C to K.
Tc = 20 + 273.15 = 293.15 K
Th = 600 + 273.15 = 873.15 K
And substitute the values into the formula.
e = 1 - 293.15/873.15
e = 1 - 0.335738418
e = 0.664261582
Since our engine is only operating at 30% of this efficiency, the actual efficiency is
0.3 * 0.664261582 = 0.199278474
So in order to do 1000 J of work, 1000 J / 0.199278474 = 5018.103448 J of energy must be extracted from the hot reservoir. Rounding to 3 significant figures gives 5020 Joules.
The energy required by the engine from the reservoir to do [tex]1000\,{\text{J}}[/tex] of work is [tex]\boxed{5020\,{\text{J}}}[/tex] .
Further Explanation:
Given:
The temperature of the first reservoir is [tex]{600\:^{\circ}\text{C}}[/tex] or [tex]873\,{\text{K}}[/tex] .
The temperature of the second reservoir is [tex]{20\:^{\circ}\text{C}}[/tex] or [tex]293\,{\text{K}}[/tex] .
Concept:
The maximum possible efficiency of a heat engine working between the energy reservoirs at different temperatures is:
[tex]\boxed{\eta=1-\frac{{{T_b}}}{{{T_a}}}}[/tex]
Here, [tex]\eta[/tex] is the efficiency of the engine, [tex]{T_a}[/tex] is the temperature of the first reservoir and [tex]{T_b}[/tex] is the temperature of the second reservoir.
Substitute the values of [tex]{T_a}[/tex] and [tex]{T_b}[/tex] in above expression.
[tex]\begin{aligned}\eta&=1-\frac{{293}}{{873}}\\&=0.664\\\end{aligned}[/tex]
Now the efficiency of the engine is [tex]30\%[/tex] of its maximum possible efficiency.
[tex]\begin{aligned}\eta'&=30\% \,{\text{of}}\,0.664\\&=\frac{{30}}{{100}}\times0.664\\&=0.199\\\end{aligned}[/tex]
The efficiency of the engine is defined as the ratio of the work done to the amount of energy extracted by the engine.
[tex]\eta '=\frac{W}{Q}[/tex]
Substitute the values of [tex]W[/tex] and [tex]\eta '[/tex] in above expression.
[tex]\begin{aligned}0.1992&=\frac{{1000\,{\text{J}}}}{Q}\\Q&=\frac{{1000}}{{0.1992}}\\&\approx5020\,{\text{J}}\\\end{aligned}[/tex]
Thus, the energy required by the engine from the reservoir to do [tex]1000\,{\text{J}}[/tex] of work is [tex]\boxed{5020\,{\text{J}}}[/tex] .
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Answer Details:
Grade: College
Subject: Physics
Chapter: Heat Engine
Keywords:
Heat engine, energy reservoirs, 30% of maximum possible efficiency, heat extracted, hot reservoir, efficiency, 20 C, 600 C, 1000 J of work.
