Question

A heat engine operating between energy reservoirs at 20∘c and 600∘c has 30% of the maximum possible efficiency. part a how much energy must this engine extract from the hot reservoir to do 1000 j of work

Answer 1

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.

Answer 2

The energy required by the engine from the reservoir to do $1000\,{\text{J}}$  of work is $\boxed{5020\,{\text{J}}}$ .

Further Explanation:

Given:

The temperature of the first reservoir is ${600\:^{\circ}\text{C}}$ or $873\,{\text{K}}$ .

The temperature of the second reservoir is ${20\:^{\circ}\text{C}}$  or $293\,{\text{K}}$ .

Concept:

The maximum possible efficiency of a heat engine working between the energy reservoirs at different temperatures is:

$\boxed{\eta=1-\frac{{{T_b}}}{{{T_a}}}}$

Here, $\eta$  is the efficiency of the engine, ${T_a}$  is the temperature of the first reservoir and ${T_b}$  is the temperature of the second reservoir.

Substitute the values of ${T_a}$  and ${T_b}$  in above expression.

$\begin{aligned}\eta&=1-\frac{{293}}{{873}}\\&=0.664\\\end{aligned}$

Now the efficiency of the engine is $30\%$  of its maximum possible efficiency.

$\begin{aligned}\eta'&=30\% \,{\text{of}}\,0.664\\&=\frac{{30}}{{100}}\times0.664\\&=0.199\\\end{aligned}$

The efficiency of the engine is defined as the ratio of the work done to the amount of energy extracted by the engine.

$\eta '=\frac{W}{Q}$

Substitute the values of $W$  and $\eta '$  in above expression.

$\begin{aligned}0.1992&=\frac{{1000\,{\text{J}}}}{Q}\\Q&=\frac{{1000}}{{0.1992}}\\&\approx5020\,{\text{J}}\\\end{aligned}$

Thus, the energy required by the engine from the reservoir to do $1000\,{\text{J}}$  of work is $\boxed{5020\,{\text{J}}}$ .

Learn More:

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2. In the calorimetry experiment which energy will be calculated during the heat exchange if water is used brainly.com/question/2566525

3. Suppose that the same amount of heat is added to two 10.0-g blocks of metal brainly.com/question/3063912

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.