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1 day 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 [3K]1 day 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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Given values:

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Solution:

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E_k = 0.5*m*v_y^2

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1 month ago

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Explanation:

a)

The car's position over time t can be described by

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This leads us to the displacement of

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The duration for this interval is

$\Delta t = 2.0 s - 0 s = 2.0 s$

Therefore, the average velocity during this period is

$v=\frac{5.6 m}{2.0 s}=2.8 m/s$

b)

At time t = 0, the position is:

$x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0$

At time t = 4.00 s, the position is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

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The position at t = 2 s is:

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And at t = 4 s it is:

$x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m$

This gives us a displacement of

$\Delta x = 20.8 - 5.6 = 15.2 m$

While the time interval is

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So the resulting average velocity is

$v=\frac{15.2}{2.0}=7.6 m/s$

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