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An ideally efficient steamboat engine operates on 520 K steam in 270 K weather. How much work can it obtain when 500 J of heat leaves the steam?

Respuesta :

opudodennis

Answer:

240 J

Explanation:

Efficiency is given by, [tex]e=\frac {T_s-T_w}{T_s}[/tex] where T is temperature and subscripts s and w to represent steam and normal weather respectively. Substituting temperature of steam and normal weather with with the given figures then efficiency will be [tex]e=\frac {520-270}{520}\approx 0.48[/tex]

Also, work done by engine is given by,

W=Qe =0.48*500=240 J

onyebuchinnaji

The work done by the steamboat when the given heat leaves it is 240 J.

The given parameters;

  • Temperature of the steam, Ts = 520 K
  • Temperature of weather, Tw = 270 K
  • heat loss, Q = 500 J

The efficiency of the steamboat is calculated as follows;

[tex]e = \frac{T_s - T_w }{T_s} \\\\e = \frac{520 - 270}{520 } \\\\e = 0.48[/tex]

The work done by the steamboat is calculated as follows;

[tex]W =Q e\\\\W = 500 \times 0.48\\\\W = 240 \ J[/tex]

Thus, the work done by the steamboat when the given heat leaves it is 240 J.

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