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3 years ago

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In Josiah Johnson Hawes and Albert Sands Southworth, Early Operation under Ether, Massachusetts General Hospital the elevated vi

ewpoint flattens the spatial perspective and emphasizes the relationships of the figures in ways the, ________________________ especially _____________, found intriguing. a. Post-Impressionists; Degas b. Impressionists; Manet c. Impressionists; Degas d. Romanticism; Renoir

1 answer:

Ronch [10]3 years ago

4 0

C and A I think cause I don’t really remember this I done before it his to be C and A

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An air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to m

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Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

$COP = \frac{T_{L}}{T_{H}-T_{L}}$

$COP = \frac{293.15\,K}{308.15\,K-293.15\,K}$

$COP = 19.543$

Lastly, the power required to operate the air conditioning system is:

$\dot W = \frac{\dot Q_{L}}{COP}$

$\dot W = \frac{32\,kW}{19.543}$

$\dot W = 1.637\,kW$

Hence, the answer is E.

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3 years ago

Read 2 more answers

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

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4 years ago