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sertanlavr [38]
4 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]4 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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A 0.20-kg stone is held 1.3 m above the top edge of a water well and then dropped into it. The well has a depth of 5.0 m. Relati
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Answer: 2.55 joules, -9.81 joules, -12.36 joules

Explanation:

the parameters given from the question are :

mass (m) = 0.20 kg

height above water (h₁) = 1.3m

depth of the well (h₂) = 5m = -5m (the negative sign is there because it is a depth below the surface )

constant value for acceleration due to gravity (g) = 9.8 m/s

  • potential energy (PE) before the stone is released = m x g x h₁

PE₁ = 0.20 x 9.8 x 1.3 = 2.55 joules

  • potential energy (PE) when it reaches the bottom of the well= m x g x h₂

 PE₂ = 0.2 x 9.8 x (-5) = -9.81 joules

  • change in potential energy = PE₂ - PE₁

  = -9.81 - 2.55 = -12.36 joules

4 0
3 years ago
A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration
Eduardwww [97]
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7 0
3 years ago
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B​
WARRIOR [948]

\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}

Explanation:

Given:

\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}

\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}

The cross product \textbf{A}×\textbf{B} is given by

\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|

=  \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}

= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}

5 0
3 years ago
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KatRina [158]

a)5m/s b)5

the 5 is because you add the seconds to get 8 seconds and then do the same with the distance to get 40. 40/8 = 5. speed = 5

Velocity = displacement/change in time

V = 40/8

I just realized how unorganised my math looks but I hope this is helpfull

3 0
3 years ago
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