How much power does a machine have that does 15204 J of work in 40 seconds?

A piston-cylinder chamber contains 0.1 m3 of 10 kg R-134a in a saturated liquid-vapor mixture state at 10 °C. It is heated at co

vaieri [72.5K]

Answer:

(A) 10132.5Pa

(B)531kJ of energy

Explanation:

This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.

Given

m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³

P1 = 101325Pa. M = 102.03g/mol

P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa

(B) Energy is transfered by the r134a in the form of thw work done in in expansion

W = nRTIn(V2/V1)

n = m / M = 10000/102.03 = 98.01mols

W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)

= 531kJ.

6 0

2 years ago

A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c

g100num [7]

Explanation:

The given data is as follows.

Mass of oxygen present = 100 mg = $100 \times 10^{-3}$ g

So, moles of oxygen present are calculated as follows.

n = $\frac{100 \times 10^{-3}}{32}$

= $3.125 \times 10^{-3}$ moles

Diameter of cylinder = 6 cm = $6 \times 10^{-2}$ m

= 0.06 m

Now, we will calculate the cross sectional area (A) as follows.

A = $\pi \times \frac{(0.06)^{2}}{4}$

= $2.82 \times 10^{-3} m^{2}$

Length of tube = 11 cm = 0.11 m

Hence, volume (V) = $2.82 \times 10^{-3} \times 0.11$

= $3.11 \times 10^{-4} m^{3}$

Now, we assume that the inside pressure is P .

And, $P_{atm}$ = 100 kPa = 100000 Pa,

Pressure difference = 100000 - P

Hence, force required to open is as follows.

Force = Pressure difference × A

= $(100000 - P) \times 2.82 \times 10^{-3}$

We are given that force is 173 N.

Thus,

$(100000 - P) \times 2.82 \times 10^{-3}$ = 173

Solving we get,

P = $3.8650 \times 10^{4} Pa$

= 38.65 kPa

According to the ideal gas equation, PV = nRT

So, we will put the values into the above formula as follows.

PV = nRT

$38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T$

T = 462.66 K

Thus, we can conclude that temperature of the gas is 462.66 K.

6 0

2 years ago

Separating the electron from the proton in a hydrogen atom takes 2.18 ✕ 10−18 j of work. through what electric potential differe

My name is Ann [436]

Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V

6 0

2 years ago

The weights of a large number of miniature poodles are approximately normally distributed with a mean of 99 kilograms and a stan

Karo-lina-s [1.5K]

Answer:

See explanation below

Explanation:

As I say in the comments, the question is incomplete, however, I will try to answer this by using data that I found on another site.

This is the part of the question that is not here:

If measurements are recorded to the nearest

tenth of a kilogram, find the fraction of these poodles

with weights

(a) over 9.5 kilograms;

(b) of at most 8.6 kilograms;

So, assuming a mean of 8 kg, and 0.9 of standard deviation, let X represents the weight of the poodles

The expression to calculate the fraction of poodle needed is:

Z = X - u / d

u: weight of the large number of poodle

d: standard deviation

Replacing data of a) wer have:

Z = 9.5 - 8 / 0.9

Z = 1.67

With this value, we need to take the value of Z, and see the area under the curve of standard deviation (see table attached)

Therefore:

P (X > 9.5) = P(Z > 1.67) = 0.5 - P (Z < 1.67) = 0.5 - 0.4525 = 0.0475

b) In this part, is the same as part a) so:

Z = 8.6 - 8 / 0.9 = 0.67

The value for area in the curve is 0.2486 so:

P = 0.5 + 0.2486 = 0.7486

Hope this helps

8 0

2 years ago

What is the value of work done on an object when a 0.1x102-newton force moves it 30 meters and the angle between the force and t

disa [49]

Answer:306joules

Explanation:

3 0

2 years ago