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lawyer [7]
3 years ago
7

A closed system contains propane at 35°c. It produces 35 kW of work while absorbing 35 kW of heat. What is process? the temperat

ure of the system after this process.
Engineering
1 answer:
salantis [7]3 years ago
3 0

Answer:

35°c

Explanation:

Given data in question

heat = 35 kw

work = 35 kw

temperature = 35°c

To find out

temperature of the system after this process

Solution

we know that first law of thermodynamics is Law of Conservation of Energy

i.e  energy can neither be created nor destroyed and it can be transferred from one form to another form

first law of thermodynamics is energy (∆E) is sum of heat (q) and work (w)

here we know

35 = 35 + m Cv ( T - t )

35-35 = m Cv ( T-t )

T = t

here T = final temperature

t = initial temperature

it show final temp is equal to initial temp

so we can say temp after process is 35°c

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We can see the pic shown in order to understand the question.

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M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\  x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m

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If M = Mmax, we have

(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4}   }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}

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