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Zinaida [17]
3 years ago
8

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

You set the bowl up so that as it cools to room temperature the heat first flows through a Carnot Engine. The soup has Cv= (33 J/K). Assume that the volume of the soup does not change. What fraction of the total heat QH that is lost by the soup can be turned into useable work by the engine?
Physics
1 answer:
DiKsa [7]3 years ago
4 0

Answer:0.061

Explanation:

Given

T_C=300 k

Temperature of soup T_H=340 K

heat capacity of soup c_v=33 J/K

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any  instant

efficiency is given by

\eta =\frac{dW}{Q}=1-\frac{T_C}{T}

dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

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mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi
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Answer:

The answer is

"x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs \frac{8}{3} feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\

The source of the hooks law is stable,

16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\

Number \frac{1}{2}  times the immediate speed, i.e .. Damping force

\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2}  \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\

The m^2+m+12=0 and m is an auxiliary equation,

m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \  m2 =\frac{-1 - \sqrt{47i}}{2}

Therefore, additional feature

x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]

Use the form of uncertain coefficients to find a particular solution.  

Assume that solution equation,

x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\

These values are replaced by equation ( 1):

\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\

Going to compare cos3 t and sin 3 t coefficients from both sides,  

The cost3 t is 3A + 3B= 20 coefficients  

The sin 3 t is 3B -3A = 0 coefficient  

The two equations solved:

3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\

Replace the very first equation with the meaning,

3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\

equation is

x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)

The ultimate plan for both the equation is therefore

x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.  

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\

x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t)  +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\

c_2=\frac{-64\sqrt{47}}{141}

x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))

5 0
3 years ago
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