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Vesna [10]
2 years ago
9

A student claims that when two bodies not initially in thermal equilibrium are placed in contact, the rise in temperature of the

cooler body must always be equal to the drop in temperature of the warmer body. Do you agree? Is there a principle of conservation of temperature or something like that?
Please answer in brief its for 10 marks​
Physics
1 answer:
zimovet [89]2 years ago
3 0

Answer:

Explanation:

No.

There is a difference between energy, called heat in this case, and temperature, which is a measure of the amount of heat contained in a material and is dependent on the material properties.

Temperature difference is what causes heat to move from one body to another.

Two objects at different temperatures placed in contact with one another will cause heat to move from the warmer body to the colder body until the temperature difference is eliminated.

The amount of heat leaving the warmer body will exactly equal the amount of heat absorbed by the cooler body. (assuming isolated system of two bodies) The temperature change within each of those bodies could be vastly different.

Example would be a 2 mm bead of molten lead dropped into a liter glass of tap water. The lead may cool several hundred °C as it solidifies while the water temperature would increase less than 1 °C

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Answer:

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Explanation:

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v=\dfrac{1}{\sqrt{k\mu_o\epsilon_o}}

Where

k is the dielectric constant of the substance.

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If you have 100 W expended over 20 s how much energy did it take?
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How long did Skylab orbit the earth?
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A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

8 0
3 years ago
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