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ivanzaharov [21]
3 years ago
7

As part of a chemistry experiment, Austin ends up with a beaker containing a mixture of two liquids, which he knows have boiling

points of 77 °C and 103 °C. Austin is already familiar with___________ as a good method for separating liquids, so he asks his chemistry teacher where to find the heating mantle and condensing tube he'll need to___________ the mixture.
Engineering
1 answer:
asambeis [7]3 years ago
7 0

Answer:

Distillation, heat

Explanation:

Here in this question, we simply want to look at the best options that could fit in the gaps.

We have a mixture of liquids having boiling points which is far from each other.

Whenever we have a mixture of liquids with boiling points far away from each other, the best technique to use in separating them is to use distillation. That is why we have that as the best fit for the first missing gap.

Now, to get the liquids to separate from each other, we shall be needing the heating mantle for the application of heat. This ensures that the mixture is vaporized. After vaporization, the condensing tube will help to condense the vapor of each of the liquids once we reach the boiling point of either of the two.

Kindly note that the liquid with the lower temperature will evaporate first and will be first obtained. In fact after reaching a little above the boiling point of the lower boiling liquid, we can be sure that what we have left in the mixture pot is the second other liquid with the higher boiling point.

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A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos
Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0
3 years ago
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emf generated by the coil is 1.57 V

Explanation:

Given details-

Number of turns of wire- 1000 turns

The diameter of the wire coil- 1 cm

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Magnetic Field (Final)=0.30 T

Time=10 ms

The orientation of the axis of the coil= parallel to the field.

We know that EMF of the coil is mathematically represented as –

E=N(ΔФ/Δt)

Where E= emf generated

ΔФ= change inmagnetic flux

Δt= change in time

N= no of turns*area of the coil

Substituting the values of the above variables

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=.0785

E=0.0785(.2/10*10-3)

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Thus, the emf generated is 1.57 V

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Answer:

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ikadub [295]

Answer:

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