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Marat540 [252]
3 years ago
10

A heat pump receives heat from a lake that has an average winter time temperature of 6o C and supplies heat into a house having

an average temperature of 25o C. (a) If the house loses heat to the atmosphere at the rate of 60,000 kJ/h, determine the minimum power supplied to the heat pump (in kW) that can maintain the interior temperature of the house at 25o C. (b) Suppose the heat pump absorbs energy from the ground (10o C) instead of the lake, Calculate the minimum power (in kW) required for the heat pump.
Physics
1 answer:
Natalija [7]3 years ago
4 0

Answer:

Explanation:

60000 kJ / h = 60000 x 1000 / (60 x 60 )

= 16667 J /s

a)

\frac{Q_H}{W} =\frac{T_H}{T_H-T_C}

where Q_H and W are heat supplied and heat given from  outside source .

Here Q_H = 16667 J/s

\frac{T_H}{T_H-T_C}=\frac{298}{298-279}

= 15.684

Q_H = 16667 J/s

16667 / W = 15.684

W= minimum power supplied  = 1062.7 W.

b )

If  T_C=283K

\frac{T_H}{T_H-T_C}=\frac{298}{298-283}

=19.87

\frac{Q_H}{W} = 19.87

W=\frac{Q_H}{19.87}

W=\frac{16667}{19.87}

= 838.8 W .

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Physical Science, please explain!
DedPeter [7]

Answer:

4N

Explanation:

because the net force is greater in the right direction

3 0
2 years ago
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Ede4ka [16]
I’m not sure if this will help but I found: https://prezi.com/l0fa6du3b9kp/going-off-the-grid-assignment/?fallback=1 and
3 0
2 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
A certain kind of glass has an index of refraction of 1.640 for blue light of wavelength 440 nm and an index of 1.595 for red li
seraphim [82]

Answer: 18.27°

Explanation:

Given

Index of refraction of blue light, n(b) = 1.64

Wavelength of blue light, λ(b) = 440 nm

Index of refraction of red light, n(r) = 1.595

Wavelength of red light, λ(r) = 670 nm

Angle of incident, θ = 30°

Angle of refraction of red light is

θ(r) = sin^-1 [(n(a)* sin θ) / n(r)], where n(a) = index of refraction of air = 1

So that,

θ(r) = sin^-1 [(1 * sin 30) / 1.595]

θ(r) = sin^-1 (0.5 / 1.595)

θ(r) = sin^-1 0.3135

θ(r) = 18.27°

4 0
2 years ago
A 5.00 kg mass is placed on top of a vertical spring, which compresses a distance of 3.13 cm. Calculate the force constant (in N
77julia77 [94]

Answer:

<h2>1567.09 N/m</h2>

Explanation:

Step one:

given data

mass m=5kg

compression x= 3.13cm to m= 0.0313m

<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>

F=ke

where

k= spring constant in N/m

e= extension/compression in

Step two:

assume g= 9.81m/s^2

F=mg

F=5*9.81

F=49.05N

substitute in the expression F=ke

49.05=k*0.0313

k=49.05/0.0313

k=1567.09 N/m

<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>

8 0
2 years ago
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