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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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Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
4 years ago
What is the formula for work?
babunello [35]
W = F • d • cos(theta)

6 0
3 years ago
Which states of matter are significantly compressible?
andreev551 [17]
Hello.


The gaseous state is the more compressible state, because it has the volume of its container.

The liquid state is virtually incompressible, and the solid state compression is very small.

The plasma is another state that has high compression, but in this case the matter is not bound(we don't have the proton in the core of the atom)
5 0
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A substance that produces H+ ion in a solution is an (n)
SpyIntel [72]

Answer:

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3 0
4 years ago
A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?
posledela

Answer:

118\; \rm Hz.

Explanation:

The frequency f of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of v= 295\; \rm m\cdot s^{-1}. In other words, the wave would have traveled 295\; \rm m in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that 295\; \rm m? The wavelength of this wave\lambda = 2.50\; \rm m gives the length of one wave cycle. Therefore:

\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118.

That is: there are 118 wave cycles in 295\; \rm m of this wave.

On the other hand, Because that 295\; \rm m of this wave goes through that point in each second, that 118 wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

f = 118\; \rm s^{-1} = 118\; \rm Hz.

The calculations above can be expressed with the formula:

\displaystyle f = \frac{v}{\lambda},

where

  • v represents the speed of this wave, and
  • \lambda represents the wavelength of this wave.

6 0
3 years ago
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