
A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated \[\left( P \right)\] by the metal. The sensor has a scale that displays \[{\log _2}\dfrac{P}{{{P_0}}}\], where \[{P_0}\] is a constant. When the metal surface is at a temperature of\[{487^\circ }C\], the sensor shows a value \[1\]. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to \[{2767^\circ }C\] ?
Answer
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Hint: In order to solve this question, we are going to first convert the two temperatures given in degree Celsius to kelvin, then, after that, from the pressure formula, the expression for \[{\log _2}\dfrac{P}{{{P_0}}}\] is found, after which the two temperature values are put along with other given values.
Formula used:
The relation for the pressure is given by the formula
\[P = \sigma Ae{T^{^4}}\]
Complete step by step answer:
It is given in the question that
\[{T_1} = {487^\circ }C = 487 + 273 = 760K\]
And the second temperature to which the metal surface is raised is
\[{T_2} = {2767^\circ }C = 2767 + 273 = 3040K\]
The relation for the pressure is given by the formula
\[P = \sigma Ae{T^{4}}\]
If, \[{P_0} = \sigma Ae\]
Then, \[P = {P_0}{T^{4}}\]
Taking logarithm to the base \[2\] on the both sides,
\[{\log _2}\dfrac{P}{{{P_0}}} = 4{\log _2}\left( T \right)\]
Putting the values of the two temperatures, we get
For, \[{T_1} = 760K\]
\[1 = 4{\log _2}\left( {760} \right)\]......................(1)
Which forms the equation number (1)
For, \[{T_2} = 3040K\]
\[{\log _2}\dfrac{P}{{{P_0}}}\, = 4{\log _2}\left( {3040} \right)\]........................(2)
Which forms the equation number (2)
Subtracting the equation (1) from (2), we get
\[{\log _2}\dfrac{P}{{{P_0}}}\, - 1 = 4{\log _2}\dfrac{{3040}}{{760}}\]
Simplifying this equation, we get
${\log _2}\dfrac{P}{{{P_0}}}\, - 1 = 4{\log _2}4 \\
\Rightarrow {\log _2}\dfrac{P}{{{P_0}}}\, - 1 = 8 \\
\therefore {\log _2}\dfrac{P}{{{P_0}}} = 9 \\$
Thus, the value displayed by the sensor when the temperature of the metal surface is raised to \[{2767^\circ }C\] is equal to \[9\].
Note: It is important to note that the pressure depends upon the temperature, density and area, where the only quantity that is variable is the temperature. This gives the relation for the ease of finding the value of the quantity \[{\log _2}\dfrac{P}{{{P_0}}}\] at the other temperature, using the one given for first temperature.
Formula used:
The relation for the pressure is given by the formula
\[P = \sigma Ae{T^{^4}}\]
Complete step by step answer:
It is given in the question that
\[{T_1} = {487^\circ }C = 487 + 273 = 760K\]
And the second temperature to which the metal surface is raised is
\[{T_2} = {2767^\circ }C = 2767 + 273 = 3040K\]
The relation for the pressure is given by the formula
\[P = \sigma Ae{T^{4}}\]
If, \[{P_0} = \sigma Ae\]
Then, \[P = {P_0}{T^{4}}\]
Taking logarithm to the base \[2\] on the both sides,
\[{\log _2}\dfrac{P}{{{P_0}}} = 4{\log _2}\left( T \right)\]
Putting the values of the two temperatures, we get
For, \[{T_1} = 760K\]
\[1 = 4{\log _2}\left( {760} \right)\]......................(1)
Which forms the equation number (1)
For, \[{T_2} = 3040K\]
\[{\log _2}\dfrac{P}{{{P_0}}}\, = 4{\log _2}\left( {3040} \right)\]........................(2)
Which forms the equation number (2)
Subtracting the equation (1) from (2), we get
\[{\log _2}\dfrac{P}{{{P_0}}}\, - 1 = 4{\log _2}\dfrac{{3040}}{{760}}\]
Simplifying this equation, we get
${\log _2}\dfrac{P}{{{P_0}}}\, - 1 = 4{\log _2}4 \\
\Rightarrow {\log _2}\dfrac{P}{{{P_0}}}\, - 1 = 8 \\
\therefore {\log _2}\dfrac{P}{{{P_0}}} = 9 \\$
Thus, the value displayed by the sensor when the temperature of the metal surface is raised to \[{2767^\circ }C\] is equal to \[9\].
Note: It is important to note that the pressure depends upon the temperature, density and area, where the only quantity that is variable is the temperature. This gives the relation for the ease of finding the value of the quantity \[{\log _2}\dfrac{P}{{{P_0}}}\] at the other temperature, using the one given for first temperature.
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