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valkas [14]
3 years ago
7

The motor in a refrigerator has a power output of 294 W. If the freezing compartment is at 271 K and the outside air is at 310 K

, assuming ideal efficiency, what is the maximum amount of heat (in joules) that can be extracted from the freezing compartment in 81.6 minutes
Physics
2 answers:
Nuetrik [128]3 years ago
7 0

Answer:

Q_{L} = 11.442\times 10^{6}\,J

Explanation:

The ideal efficiency of a refrigerator is given by the model of a Carnot refrigerator, whose indicator is known as Coefficient of Performance:

COP_{R} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{R} = \frac{310\,K}{310\,K-271\,K}

COP_{R}=7.949

Heat rate extracted from the freezing compartment is:

\dot Q_{L} = COP_{R}\cdot \dot W

\dot Q_{L} = 7.949\cdot 294\,W

\dot Q_{L} = 2337.01\,W

The heat that can be extracted in abovementioned period of time is:

Q_{L} = \dot Q_{L}\cdot \Delta t

Q_{L} = (2337.01\,W)\cdot (81.6\,min)\cdot (\frac{60\,s}{1\,min} )

Q_{L} = 11.442\times 10^{6}\,J

dusya [7]3 years ago
5 0

Answer:

Maximum amount of heat = 10002151.38J

Explanation:

Workdone by motor in 86.1 minutes I given by:

W = power × time

W= 294 × 86.1×60

W= 1439424 Joules

W= 1.4 ×10^6Joules

The amount of heat extracted is given by:

/QL /= K/W/ = TL/W/ /(TH - TL)

Where TL= freezing compartment temperature

TH = Outside air temperature

/QL /= 271 × 1439424 / (310 - 271)

/QL/ = 390083904/39

/QL/ = 10002151.38 Joules

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<h2>Answer: D 60N</h2>

<h3>Explanation:</h3>

Mass(M)=15 kg

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Force=?

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Force(F)=M×A

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A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it
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Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

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Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

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1.3 = 1.8 * t1

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Answer:

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Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²

F = 2.16×10⁻⁶ N

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