5020 Joules.The maximum theoretical performance of an warmth engine ise = 1 - Tc/Thwheree = efficiencyTc = cold temperature (absolute)Th = warm temperature (absolute)So let"s transform the temperatures from C to K.Tc = 20 + 273.15 = 293.15 KTh = 600 + 273.15 = 873.15 KAnd instead of the values into the formula.e = 1 - 293.15/873.15e = 1 - 0.335738418e = 0.664261582Since our engine is only operating at 30% that this efficiency, the actual effectiveness is0.3 * 0.664261582 = 0.199278474So in order to carry out 1000 J the work, 1000 J / 0.199278474 = 5018.103448 J of power must be extract from the hot reservoir. Rounding to 3 far-reaching figures provides 5020 Joules.

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photoshop1234 <79>6 months ago
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The power required by the engine from the reservoir to execute

of work-related is
.

Further Explanation:

Given:

The temperature the the an initial reservoir is

or
.

The temperature the the second reservoir is

or
.

Concept:

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

Here,

is the effectiveness of the engine,
is the temperature of the first reservoir and
is the temperature that the second reservoir.

Substitute the values of

and also
in over expression.

Now the efficiency of the engine is

the its maximum possible efficiency.

The effectiveness of the engine is characterized as the proportion of the job-related done come the quantity of power extracted by the engine.

Substitute the values of

and
in over expression.

Thus, the energy required by the engine from the reservoir to execute

of occupational is
.

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3.

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