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Evgesh-ka [11]
3 years ago
6

Why do you think the temperature does not change much during a phase change?

Physics
2 answers:
Zielflug [23.3K]3 years ago
8 0

Answer:

<em>By </em><em>the </em><em>there </em><em>is </em><em>no </em><em>change</em><em> </em><em>in </em><em>the </em><em>temperature</em><em> </em><em>until </em><em>the </em><em>phase </em><em>change </em><em>is </em><em>complete.</em><em> </em><em>It </em><em>is </em><em>tha</em><em>t</em><em> </em><em>during</em><em> </em><em>the </em><em>phase </em><em>change </em><em>the </em><em>energy </em><em>supplied </em><em>is </em><em>used </em><em>only </em><em>to </em><em>seperate </em><em>the </em><em>molecules </em><em>no </em><em>part </em><em>of </em><em>it </em><em>is </em><em>used </em><em>to </em><em>increase </em><em>the </em><em>relating </em><em>to </em><em>a </em><em>result </em><em>from </em><em>the </em><em>motion </em><em>energy</em><em> </em><em>of </em><em>the </em><em>molecules </em><em>so </em><em>it's </em><em>temperature</em><em> </em><em>will </em><em>not </em><em>rise </em><em>since </em><em>the </em><em>relating</em><em> to</em><em> a</em><em> </em><em>result</em><em> </em><em>from</em><em> the</em><em> </em><em>motion </em><em>energy </em><em>of </em><em>molecules </em><em>remains </em><em>the </em><em>same </em><em>.</em>

<em><u>I </u></em><em><u>hope </u></em><em><u>this </u></em><em><u>might </u></em><em><u>help </u></em><em><u>u</u></em><em> </em>

11111nata11111 [884]3 years ago
3 0

Answer:

It depends on where the temperature is dropping, in which body so to speak. Generally, the temperature adapts to the two bodies, for example if a hot piece of metal meets a cold one, the two will continue until they are at an equal temperature, an intermediate temperature.

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A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that
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B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.

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Answer:

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By method of dimension show that the following equation are homogenous.
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Answer:

Proof in explanataion

Explanation:

The basic dimensions are as follows:

MASS = M

LENGTH = L

TIME = T

i)

Given equation is:

H = \frac{u^2Sin^2\phi}{2g}

where,

H = height (meters)

u = speed (m/s)

g = acceleration due to gravity (m/s²)

Sin Ф = constant (no unit)

So there dimensions will be:

H = [L]

u = [LT⁻¹]

g = [LT⁻²]

Sin Ф = no dimension

Therefore,

[L] = \frac{[LT^{-1}]^2}{[LT^{-2}]}\\\\\ [L] = [L^{(2-1)}T^{(-2+2)}]

<u>[L] = [L]</u>

Hence, the equation is proven to be homogenous.

ii)

F = \frac{Gm_1m_2}{r^2}\\\\

where,

F = Force = Newton = kg.m/s² = [MLT⁻²]

G = Gravitational Constant = N.m²/kg² = (kg.m/s²)m²/kg² = m³/kg.s²

G = [M⁻¹L³T⁻²]

m₁ = m₂ = mass = kg = [M]

r = distance = m = [L]

Therefore,

[MLT^{-2}] = \frac{[M^{-1}L^{3}T^{-2}][M][M]}{[L]^2}\\\\\ [MLT^{-2}] = [M^{(-1+1+1)}L^{(3-2)}T^{-2}]\\\\

<u>[MLT⁻²] = [MLT⁻²]</u>

Hence, the equation is proven to be homogenous.

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3 years ago
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