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3 years ago

5

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

reservoir. during that time, what is the maximum amount of work wmax that the engine might have performed?

1 answer:

tino4ka555 [31]3 years ago

6 0

$q_{c}$ = Heat released to cold reservoir

$q_{h}$ = Heat released to hot reservoir

$W_{max}$ = maximum amount of work

$t_{c}$ = temperature of cold reservoir

$t_{h}$ = temperature of hot reservoir

we know that

$\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}$

$q_{h} = (\frac{t_{h}}{t_{c}})q_{c}$ eq-1

maximum work is given as

$W_{max}$ = $q_{h}$ - $q_{c}$

using eq-1

$W_{max}$ = $(\frac{t_{h}}{t_{c}})q_{c}$ - $q_{c}$

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A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

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And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

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$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

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Complete Question

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Answer:

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Explanation:

From the question we are told that

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