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sertanlavr [38]
3 years ago
14

An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro

pped into boiling water at 97°C. Taking the properties of the egg to be r 1020 kg/m3 and cp 3.32kJ/kg · °C, determine (a) how much heat is transferred to the egg by the time the average temperature of the egg rises to70°C and (b) the amount of entropy generation associated with this heat transfer process.
Physics
1 answer:
kupik [55]3 years ago
3 0

Answer:

a) Q_{in} = 13.742\,kW, b) \Delta S = 370.15\,\frac{kJ}{K}

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

Q_{in} +U_{sys,1} - U_{sys,2} = 0

Q_{in} = U_{sys,2} - U_{sys,1}

Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})

Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)

Q_{in} = 13.742\,kW

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

\Delta S = \frac{Q_{in}}{T_{in}}

\Delta S = \frac{13.742\,kJ}{370.15\,K}

\Delta S = 370.15\,\frac{kJ}{K}

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3 years ago
Garrick rubs an inflated balloon against his hair. He then touches the balloon against a non-conducting wall.
Sauron [17]

Answer:

Figure A

Explanation:

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In this situation, the balloon is charged by friction: because of the friction between the surface of the balllon and the hair, electrons are transferred from the hair to the surface of the balloon.

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Then, the balloon is placed in contact with the non-conducting wall.

The non-conducting wall is initially neutral (equal number of positive and negative charges).

Because the wall is made of a non-conducting material (=isolant), the charges cannot move easily through it. Therefore, even though the charges on the wall feel a force due to the presence of the electrons in the balloon, they will not redistribute along the wall.

Therefore, the charges on the wall will remain equally distributed, as shown in figure A.

7 0
4 years ago
A charge of 2 c is at the origin. when charge q is placed at 2 m along the positive x axis, the electric field at 2 m along the
docker41 [41]

The value of Q will be -8 C.

In the presence of an electric or magnetic field, matter experiences a force due to its electric charge.

A moving electric charge generates a magnetic field, and an electric charge has an accompanying electric field.

The information provided in the issue is;

The separation between and is 2m.

The separation between and is 2m.

An origin charge equals +2 C

The electric fields are identical in magnitude but are facing in different directions. As a result, the following relationship can be used

Q/16=1/2

The value of Q will be -8 C.

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brainly.com/question/8163163

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4 0
2 years ago
Convert 23 miles into feet.<br> **There are 5280 feet in 1 mile** *
zhenek [66]

Answer:

121440 feet is 23 miles.

5280 x 23= 121440

To check your answer just divide 121440 by 5280, which gives you 23.

Hopefully this helps you!

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3 years ago
A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C. What is the entropy change of the sample if the latent hea
Sunny_sXe [5.5K]

Answer:

The entropy change of the sample of water =  6.059 x 10³ J/K.mol

Explanation:

Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol

Mathematically, entropy is expressed as

ΔS = ΔH/T....................... Equation 1

Where ΔH = heat absorbed or evolved, T = absolute temperature.

<em>Given:  If 1 mole of water = 0.0018 kg,</em>

<em>ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.</em>

<em>T = 100 °C = (100+273)  K = 373 K.</em>

<em>Substituting these values into equation 1,</em>

<em>ΔS =2.26x 10⁶/373</em>

ΔS = 6.059 x 10³ J/K.mol

Therefore the entropy change of the sample of water =  6.059 x 10³ J/K.mol

7 0
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