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3 years ago

For a refrigerator, COP<1 a) True b) False

Engineering

1 answer:

WINSTONCH [101]3 years ago

4 0

Answer:

False

Explanation:

COP is Coeficient of performance, it relates useful heat moved with work consumed, it is used in refrigerators and heat pumps.

For refrigerators the useful heat is the one extracted from the cold reservoir. Therefore:

$COP = \frac{Tcold}{T.hot - Tcold}$

This often exceeds 1.

You might be interested in

one number is 11 more than another number. find the two number if three times the larger number exceeds four times the smaller n

vaieri [72.5K]

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

$\begin{bmatrix}1 & -1 & 11\\3 & -4 & 4 \end{bmatrix}$

$\begin{bmatrix}1 & 0 & 40\\0 & 1 & 29 \end{bmatrix}$

a = 40

b = 29

6 0

3 years ago

A piston–cylinder device contains a mixture of 0.5 kg of H2 and 1.2 kg of N2 at 100 kPa and 300 K. Heat is now transferred to th

Taya2010 [7]

Answer:

(a) The heat transferred is 2552.64 kJ

(b) The entropy change of the mixture is 1066.0279 J/K

Explanation:

Here we have

Molar mass of H₂ = 2.01588 g/mol

Molar mass of N₂ = 28.0134 g/mol

Number of moles of H₂ = 500/2.01588 = 248 moles

Number of moles of N₂ = 1200/28.0134 = 42.8 moles

P·V = n·R·T

V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³

Since the volume is doubled then

V₂ = 2 × 7.25 = 14.51 m³

At constant pressure, the temperature is doubled, therefore

T₂ = 600 K

If we assume constant specific heat at the average temperature, we have

Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂

cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K

cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K

Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ

b) $\Delta S = - R(n_A \times lnx_A + n_B \times ln x_B)$

Where:

$x_A$ and $x_B$ are the mole fractions of Hydrogen and nitrogen respectively.

Therefore, $x_A$ = 248 /(248 + 42.8) = 0.83

$x_B$ = 42.8/(248 + 42.8) = 0.1472

∴ $\Delta S = - 8.3145(248 \times ln0.83 + 42.8 \times ln 0.1472)$ = 1066.0279 J/K

5 0

3 years ago

The following laboratory test results for Atterberg limits and sieve-analysis were obtained for an inorganic soil. [6 points] Si

alexira [117]

Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

<u>A) Classifying the soil according to USCS system</u>

( using 2nd image attached below )

<em>description of sand</em> :

The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

5 0

3 years ago

A metal bar has a 0.6 in. x 0.6 in. cross section and a gauge length of 2 in. The bar is loaded with a tensile force of 50,000 l

Aleks [24]

Answer:

modulus =3.97X10^6 Ib/in^2, Poisson's ratio = 0.048

Explanation:

Modulus is the ratio of tensile stress to tensile strain

Poisson's ratio is the ratio of transverse contraction strain to longitudinal extension strain within the direction of the stretching force

And contraction occur from 0.6 in x 0.6 in to 0.599 in x 0.599 in while 2 in extended to 2.007, with extension of 0.007 in

5 0

3 years ago

Which of the following is an example of a hardwood? A maple B spruce C pine D fir

bearhunter [10]

Answer:

A. Maple

Explanation:

Maple is a hardwood.

Hope that helps!

7 0

2 years ago