Physics: A substance has a mass of 0.02kg and a volume of 0.08m. What is its density? Give your answer to one decimal place A substance has a mass of 0.02kg and a volume of 0.08m3. What is its density? Give your answer to one decimal place.

The emissivity of the radiation shield is Explanation:From the question we are told that The temperature of the first parallel plate is The temperature of the second parallel plate is The emissivity of first plate is The emissivity of first plate is Generally the total radiation heat that is been transferred without the shield is mathematically represented as Where is the Stefan-Boltzmann constant which has a value Substituting values ...

A substance has a mass of 0.02kg and a volume of

Ответ:

tiakimberly3

04.10.2020

Physics

The emissivity of the radiation shield is e_3 = 0.580

Explanation:

From the question we are told that

The temperature of the first parallel plate is T_1 = 650K

The temperature of the second parallel plate is T_2 = 400K

The emissivity of first plate is e_1 = 0.6

The emissivity of first plate is e_2 = 0.9

Generally the total radiation heat that is been transferred without the shield is mathematically represented as

$Q_1 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{\frac{1}{e1 } + \frac{1}{e_2} -1 }$

Where $\sigma$ is the Stefan-Boltzmann constant which has a value $5.67 *10^{-8} \ W \cdot m^{-2} \cdot K^{-1}$

Substituting values

$Q_1 = \frac{ 5.67 *10^{-8} (650 ^4 - 400 ^4)}{\frac{1}{0.6 } + \frac{1}{0.9} -1 }$

$Q_1 = 4876.8 \ W/m^2$

From the question we are told the that using the radiation shield would reduce the radiation heat transfer by 15%

So the new heat transfer is

$Q_2 = \frac{15}{100} * Q_1$

So $Q_2 = \frac{15}{100} * 4876.8$

Q_2 = 731.52 W/m^2

Now this new radiation heat transfer can be mathematically represented as

$Q_2 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{ [\frac{1}{e_1 } + \frac{1}{e_2 } - 1 ] + n [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

Where e_3 the emissivity of the radiation shield and n is the number of radiation shield

Substituting values

$731.52 = \frac{5.67 *10^{-8} (650 ^4 - 400 ^4)}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

$731.52 = \frac{1.4175*10^{-5}}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

$[\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = \frac{1.4175*10^{-5}}{731.52}$

$[\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8}$

$1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8} - [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ]$

$\frac{1}{e_3} + \frac{1}{e_3} = 1 .7222222416$

e_3 = 0.580

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A substance has a mass of 0.02kg and a volume of 0.08m. What is its density? Give your answer to one decimal place

A substance has a mass of 0.02kg and a volume of 0.08m3. What is its density? Give your answer to one decimal place.

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