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alekssr [168]
3 years ago
11

Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms

, 314 K. The room serviced by unit A is kept at a temperature of 292 K, while the room serviced by unit B is kept at 298 K. The heat removed from either room is 4430 J. For both units, find (a), (b) the magnitude of the work required and (c), (d) the magnitude of the heat deposited outside for A and B conditioners respectively
Physics
1 answer:
lakkis [162]3 years ago
4 0

Answer:

(a)  333.77 J

(b) 237.85 J

(c) 4763.77 J

(d) 4667.85 J

Explanation:

Temperature of source, TH = 314 K

Temperature of A, Tc = 292 K

Temperature of B, Tc' = 298 K

heat taken out, Qc = 4430 J

Let the heat deposited outside is QH and QH' by A and B respectively.

\frac{Q_H}{Q_c}=\frac{T_H}{T_c}\\\\Q_H = \frac{4430\times314}{292}=4763.77 J

Now

\frac{Q_H'}{Q_c}=\frac{T_H}{T_c'}\\\\Q_H' = \frac{4430\times314}{298}=4667.85 J

(a) Work done for A

W = QH - QC = 4763.77 - 4430 = 333.77 J

(b) Work done for B

W' = QH' - Qc = 4667.85 - 4430 = 237.85 J

(c) QH = 4763.77 J

(d) QH' = 4667.85 J

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Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4
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Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

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Donde f es la frecuencia, en hertz.

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T = \frac{1}{4\,hz}

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Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

F \approx 80.852\,N

La velocidad lineal de la lavadora es:

v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

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