Ask question

Ask question

All categories

marishachu [46]

3 years ago

9

A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is us

ed to heat a home to 24.0°C during the winter with the low temperature reservoir at the outdoor temperature. At which outdoor temperature would it be more efficient to add the energy directly to the interior of the home than use it to run the heat pump?

1 answer:

Nataliya [291]3 years ago

6 0

Answer:

$T_C$ =118.8 K= 154.2°C

Explanation:

COP_max of carnot heat pump= $\frac{T_{H} }{T_{H}-T_{C} }$

where T_H and T_C are temperatures of hot and cold reservoirs

Also COP= $\frac{Q_H}{W}$

in the question $COP= \frac{60}{100} \times COP_{max}$

⇒ $\frac{Q_H}{W} =\frac{60}{100}\times\frac{T_H}{T_H-T_C}$

heat is added directly to be as efficient as via heat pump

$Q_H= W$

and T_H= 24° C= 297 K

$1=\frac{60}{100}\times \frac{297}{297-T_C}$

on calculating the above equation we get

$T_C$ =118.8 K

the outdoor temperature for efficient addition of heat to interior of home

$T_C$ =118.8 K= 154.2°C

You might be interested in

Difference between derived quantity and physical quantity

Lerok [7]

Answer:

<h2>Derived quantities are based on fundamental quantities, and they can be given in terms of fundamental quantities.</h2>

<h3>Fundamental quantities are the base quantities of a unit system, and they are defined independent of the other quantities. </h3>

Explanation:

#Let's Study

#I Hope It's Helps

#Keep On Learning

#Carry On Learning

$\\ \blue{MaggieEve}$

6 0

2 years ago

Me ajudem, Por favor!!!!!!

zzz [600]

Answer:

a) $a=4\,\frac{m}{s^2}$

b) $V(t)=4\,t\,+3$

c) $V(1)=7 \,\frac{m}{s} \\$

d) Displacement = 22 m

e) Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

$slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}$

Therefore, acceleration is $a=4\,\frac{m}{s^2}$

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

$y=m\,x+b\\V(t)=4\,t\,+3$

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

$V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}$

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = $\frac{(7+15)\,2}{2} = 22\,\,m$

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = $\frac{22}{2} = 11\, \,\frac{m}{s}$

3 0

3 years ago

Which region of the early universe was most likely to become a galaxy?

kramer

Answer:

This is likely possible for a region whose matter density is higher than the normal average.

Explanation:

A galaxy is a collection of lumps in space which are clumped together and interact with each other. There are a lot of speculations on how galaxies were birthed. some believe its formed by a collection of massive gas, dust which eventually collapsed under their own gravitational pull. others says its formed by the combination of large lumps of matter which accumulated forming thee galaxies. The possibility of a galaxy forming is dependent on how massive the matter in the region of the universe is.

3 0

3 years ago

An electron has a charge of 1.602 X 10-19.coulomb. When two electrons are separated by 1.2 X 10-9m, what force will they exert o

amid [387]

Answer:

The force they will exert on each other is 1.6*10⁻¹⁰ N

Explanation:

The electromagnetic force is the interaction that occurs between bodies that have an electric charge. When the charges are at rest, the interaction between them is called the electrostatic force. Depending on the sign of the interacting charges, the electrostatic force can be attractive or repulsive. The electrostatic interaction between charges of the same sign is repulsive, while the interaction between charges of the opposite sign is attractive.

Coulomb's law is used to calculate the electric force acting between two charges at rest. This force depends on the distance "r" between the electrons and the charge of both.

Coulomb's law is represented by:

$F=k*\frac{q1*q2}{r^{2} }$

where:

F = electric force of attraction or repulsion in Newtons (N). Like charges repel and opposite charges attract.
k = is the Coulomb constant or electrical constant of proportionality.
q = value of the electric charges measured in Coulomb (C).
r = distance that separates the charges and that is measured in meters (m).

In this case:

k= 9*10⁹ $\frac{N*m^{2} }{C^{2} }$
q1= 1.602*10⁻¹⁹ C
q2= 1.602*10⁻¹⁹ C
r= 1.2*10⁻⁹ m

Replacing:

$F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{1.602*10^{-19} C*1.602*10^{-19} C}{(1.2*10^{-9} )^{2} }$

and solving you get:

F=1.6*10⁻¹⁰ N

<u><em>The force they will exert on each other is 1.6*10⁻¹⁰ N</em></u>

3 0

3 years ago

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

kramer

Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass $M_{Io}=8.93\times10^{22}\ kg$

Radius = 1821 km

Height $h_{Io}=500\ km$

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

$g =\dfrac{GM_{Io}}{(R_{Io})^2}$

Put the value into the formula

$g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}$

$g=1.79\ m/s^2$

Let v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

$H=\dfrac{v^2}{2g}$

Using ratio of height of earth and height of Io

$\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}$

$\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}$

Put the value into the formula

$\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}$

$\dfrac{H_{e}}{H_{Io}}=0.182$

$H_{e}=0.182\times H_{Io}$

$H_{e}=0.182\times500$

$H_{e}=91\ km$

Hence, The height reached by the material on Earth is 91 km.

3 0

3 years ago