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Question: If an object at absolute temperature (T) radiates energy at rate \(R\), then select the correct grap...

If an object at absolute temperature (T) radiates energy at rate RR, then select the correct graph showing the variation of log0R{\log _0}R with log0(T){\log _0}(T).
A)

B)

C)

D)

Explanation

Solution

Rate of radiation of energy is known as power. We need to know the relation between rate of energy radiated and absolute temperature. Also understand what is meant by absolute temperature. Remember that absolute temperature and absolute zero temperature are related to each other but they are not the same.

Complete step by step solution:
When we start measuring temperature as 0K0K at the lowest possible energy state then temperature measured with respect to this zero is known as the absolute temperature of the object.
This measured 0K0K is known as the absolute zero temperature of the object.
Stefan-Boltzmann law gives us the relation between the power radiated and the absolute temperature of the object.
The law states that the power radiated by the object is directly proportional to the fourth power of the absolute temperature.
PT4P \propto {T^4}
Mathematically, the equation is given as
P=εσAT4P = \varepsilon \sigma A{T^4}
Where PP is the power radiated
ε\varepsilon is the emissivity
σ\sigma is the Stefan’s constant
AA is the radiating area
TT is the absolute temperature
Here in this question, the power radiated or the rate of energy radiated is represented as RR .
R=εσAT4\Rightarrow R = \varepsilon \sigma A{T^4}
Taking log0{\log _0} on both sides of the equation, we get
log0R=log0(εσAT4)\Rightarrow {\log _0}R = {\log _0}(\varepsilon \sigma A{T^4})
log0R=log0ε+log0σ+log0A+log0T4\Rightarrow {\log _0}R = {\log _0}\varepsilon + {\log _0}\sigma + {\log _0}A + {\log _0}{T^4}
ε,σ,A\because \varepsilon ,\sigma ,A are all constants log0{\log _0} of all constants is also constant.
Therefore we can substitute log0ε+log0σ+log0A=k{\log _0}\varepsilon + {\log _0}\sigma + {\log _0}A = k
log0R=k+log0T4\Rightarrow {\log _0}R = k + {\log _0}{T^4}
We can represent this equation as a linear graph of log0R{\log _0}R VS log0T{\log _0}T with kk as the intercept.

Therefore, option (A)(A) is the correct graph where log0R{\log _0}R starts with an initial value of kk and then increases linearly with increase in the value of log0T{\log _0}T .

Note: Stefan’s constant σ=5.670×108watts/m2K4\sigma = 5.670 \times {10^8}watts/{m^2}{K^4} . It is the constant of proportionality in the Stefan-Boltzmann law. The emissivity ε=1\varepsilon = 1 for black body and ε<1\varepsilon < 1 for grey bodies. Stefan-Boltzmann law is used to calculate the temperature of the Sun.