Ask question

Ask question

All categories

3 years ago

12

An air conditioner is a refrigerator with the inside of the house acting as the cold reservoir and the outside atmosphere acting

as the hot reservoir. Assume that an air conditioner consumes 1.25 * 103 W of electrical power and that it can be idealized as a reversible Carnot refrigerator. If the coefficient of performance of this device is 1.75, how much heat can be extracted from the house in a day

1 answer:

Monica [59]3 years ago

8 0

Answer:

1.89*10^8 J

Explanation:

The coefficient of performance of this device is $\frac{Q}{W}$ where Q .is the useful heat supplied or removed by the considered system and W is the work required by the considered system.

Step 1

Coeffiecient of perfromance for cooling (COPC) = 1.75 = $\frac{Q}{W}$ .

Q = 1.75W

Step 2

We convert day into seconds:

1 day = 24 hrs = 86400 seconds

Step 3

Heat that can be extracted from the house in a day is:

Q = 1.75 * 1.25 * 10^3 * 86400 = 189000000 = 1.89*10^8 J

You might be interested in

A portable CD player uses two 1.5 volt batteries. if the current in the CD player is 2amps ,what is resistance?

chubhunter [2.5K]

resistance=voltage/current

1.5/2=0.75ohms

7 0

2 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

Cerrena [4.2K]

Answer:

B false

Explanation:

dont have any explanation

8 0

2 years ago

Read 2 more answers

How can the rate of a reaction be decreased?

andre [41]

Answer:

Option A

Lowering the amount of reactants

Explanation:

To reduce the rate of chemical reaction, one can reduce temperature or surface area. The addition of catalysts increases rate of reaction but decreasing the amount of reactants decreases rate of reaction. Therefore, from the choices provided, choice A is correct.

6 0

2 years ago

The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li

melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate $Q$ of a hypodermic needle:

$Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}$ (1)

Where the dimensions of each one is:

Volume flow rate $Q=\frac{L^{3}}{T}$

Radius of the needle $R=L$

Length of the needle $L=L$

Pressures at opposite ends of the needle $P_{2}$ and $P_{1}=\frac{M}{LT^{2}}$

Viscosity of the liquid $\eta=\frac{M}{LT}$

We need to find the value of $n$ whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

$\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}$ (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}$ (3)

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}$ (4)

As we can see $n$ must be 4 if we want the exponent to be 3:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}$ (5)

Finally:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}$ (6)

8 0

3 years ago

mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi

valina [46]

Answer:

The answer is

" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

$W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\$

The source of the hooks law is stable,

$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

Assume that solution equation,

$x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\$

These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

Going to compare cos3 t and sin 3 t coefficients from both sides,

The cost3 t is 3A + 3B= 20 coefficients

The sin 3 t is 3B -3A = 0 coefficient

The two equations solved:

$3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\$

Replace the very first equation with the meaning,

$3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\$

equation is

$x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)$

The ultimate plan for both the equation is therefore

$x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)$

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

$x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\$

$x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\$

$c_2=\frac{-64\sqrt{47}}{141}$

$x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$

5 0

2 years ago