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1 month ago

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2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initia

l temperature at 1336K. The solid and liquid can only exchange heat with each other. What kind of analysis do you need to perform in order to determine whether, once thermal equilibrium is reached, the mixture will be entirely solid or in a mixed solid/liquid phase?

1 answer:

Sav [3.1K]1 month ago

8 0

Response:

Clarification:

Gold has a specific heat of 129 J/kg°C

with a melting temperature of 1336 K

and a heat of fusion of 63000 J/kg

To ensure the gold solidifies, the thermal energy required must be lower than that required to reach its melting point. Thus,

The first energy calculation is E1 = 63000 J/kg × 1.5 = 94500 J

while the second energy calculation is E2 = 129 J/kg°C × 2 kg × (1336–1000) K = 86688 J

Consequently, asserting that all is solid is incorrect. There will be a combination of solid and liquid.

For further details, the difference between E1 and E2 is 7812 J, capable of melting

7812/63000 = 0.124 kg of the solid gold

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1 month ago

Read 2 more answers

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Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Utilizing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Now let’s solve for t.

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

The computed velocity magnitude is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The ball's angle is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

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3 months ago

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Keith_Richards [3271]

Result:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The locations likely to receive snowfall due to lake effect include Grand Marais, Two Harbors, and Duluth. These areas are situated directly along the shores of Lake Superior, one of the world’s largest lakes. Its substantial water volume significantly influences the local climate, generating a lot of humidity in the air and considerable evaporation, both of which lead to cloud formation and, when temperatures drop adequately, result in significant snowfall rather than rain.

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2 months ago

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

serg [3582]

Answer:

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Explanation:

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d = Distance = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

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$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference across the plates amounts to 10000 V

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The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

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Refer to the attached file for the solution

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