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Nonamiya [84]
3 years ago
10

An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

ch of that delivered heat was originally work consumed in the transfer?
Physics
1 answer:
choli [55]3 years ago
5 0

Answer:

Wnet, in, = 133.33J

Explanation:

Given that

Pump heat QH = 1000J

Warm temperature TH= 300K

Cold temperature TL= 260K

Since the heat pump is completely reversible, the combination of coefficient of performance expression is given as,

From first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power required to drive the the heat pump is given as

Wnet, in= QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

So the 133.33J was the amount heat that was originally work consumed in the transfer.

Extra....

According to the first law, the rate at which heat is removed from the low temperature reservoir is given as

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

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Propose a hypothesis for how the position of the ball will affect the amount of its gravitational pull energy
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4 0
3 years ago
QUESTION 1
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Question 1: C Question 2: B, Hope this Helps!
3 0
3 years ago
If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t
Aleksandr-060686 [28]

Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

B = \frac{\mu I}{2 \pi r}

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

Let \ \frac{\mu I}{2\pi}  \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

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3 years ago
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