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Degger [83]
4 years ago
5

A cup of hot coffee can be cooled by placing a cold spoon in it. A spoon of which of the following materials would be most effec

tive for this purpose: aluminum, copper, iron, or glass? Assume all spoons have the same mass.
Physics
1 answer:
arsen [322]4 years ago
8 0

Answer: copper

Explanation: when a spoon is placed in a hot coffee, the coffee will cool off based on the concept of heat transfer.

The coffee is at a higher temperature compared to the spoon, hence there is conduction of heat by the spoon from the coffee.

Heat is now transfered from the coffee to the spoon, the rate of heat transfer ( for this case coffee to spoon) is dependent on some properties but for this case, the thermal conductivity is of the most important.

A material of high thermal conductivity have a very fast rate of heat transfer and a material of low thermal conductivity have a slow rate of heat transfer.

Below is the value of thermal conductivity for copper, aluminum, iron and glass.

Copper (385 w/m.K)

Aluminum (205 W/m.K)

Iron (79.5 W/m.K)

Glass (0.8 W/m.K)

It can been seen that copper has the highest value for thermal conductivity which implies that it will conduct heat faster compared to others thus cooling the coffee faster.

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3 years ago
Two very small charged particles exert an electrostatic force F on each other. If the distance between them is doubled, the forc
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3 years ago
An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
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a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

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a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

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let's locate a reference system in the load that is on the left side, the distances are

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right side -electron     r₂ = d-x

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we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

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now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

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0.05        27.0 10-16

0.10          8.10 10-16

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0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

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