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3 years ago

How does vacuum reduce thermal energy transfer

Physics

2 answers:

Evgesh-ka [11]3 years ago

4 0

Answer:

Less air pressure.

Explanation:

Thermal energy travels in many different ways and one of those ways is through the air so if there's less air, it would reduce thermal energy transfer.

FrozenT [24]3 years ago

3 0

The silver coating on the inner bottle prevents heat transfer by radiation, and the vacuum between its double wall prevents heat moving by convection. The thinness of the glass walls stops heat entering or leaving the flask by conduction.

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Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago

Suppose you take

kati45 [8]

Answer:

a current will be induced.

Explanation:

4 0

3 years ago

Read 2 more answers

During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a vel

zmey [24]

Answer:

Mass of receiver is 92 kg

Explanation:

We have given mass of tackler $m_1=100kg$

Let the mass of receiver is $m_2kg$

When tackler moving alone velocity is $v_i=4.8$ m/sec

And when tackler and receiver is together velocity is $v_f$ = 2.5 m/sec

So from conservation of momentum

$m_1v_i=(m_1+m_2)v_f$

$100\times 4.8=(100+m_2)\times 2.5$

$2.5m_2+250=480$

$2.5m_2=230$

$m_2=92kg$

So mass of receiver is 92 kg

4 0

3 years ago

When you speak or sing, you force air through your ?

zaharov [31]

You force air from your lungs up through your larynx which is commonly called the voicebox

6 0

3 years ago

Can I please have help

NeTakaya

Answer:

it will usually increase

Explanation:

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6 0

2 years ago