Which of the following graphs correctly represents the relation between In (E) and In (T), where E is the amount of radiation emitted per unit time from a unit area of the body and T is the absolute temperature ?

Options

A.
B.
C.
D.

Answer

To determine the correct graph representing the relationship between \( \ln(E) \) and \( \ln(T) \), where \( E \) is the amount of radiation emitted per unit time from a unit area of the body and \( T \) is the absolute temperature, we can follow these steps:

### Step 1: Understand Stefan-Boltzmann Law
According to the Stefan-Boltzmann law, the amount of radiation emitted per unit area \( E \) is proportional to the fourth power of the absolute temperature \( T \):
\[ E = \sigma T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant.

### Step 2: Take the Natural Logarithm
To find the relationship between \( \ln(E) \) and \( \ln(T) \), we take the natural logarithm of both sides of the equation:
\[ \ln(E) = \ln(\sigma T^4) \]

### Step 3: Apply Logarithmic Properties
Using the properties of logarithms, we can expand the right-hand side:
\[ \ln(E) = \ln(\sigma) + \ln(T^4) \]
\[ \ln(E) = \ln(\sigma) + 4\ln(T) \]

### Step 4: Rearrange the Equation
We can rearrange the equation to express it in the form of \( y = mx + c \):
\[ \ln(E) = 4\ln(T) + \ln(\sigma) \]
This indicates that \( \ln(E) \) is a linear function of \( \ln(T) \) with a slope of 4 and a y-intercept of \( \ln(\sigma) \).

### Step 5: Analyze the Graph
Since \( \sigma \) (the Stefan-Boltzmann constant) is a positive constant, \( \ln(\sigma) \) will be a negative value (as \( \sigma < 1 \)). Therefore, the y-intercept of the graph will be negative. The slope of 4 indicates that for every unit increase in \( \ln(T) \), \( \ln(E) \) increases by 4 units.

### Step 6: Conclusion
The graph that correctly represents this relationship will have a positive slope (indicating that as \( T \) increases, \( E \) also increases) and a negative y-intercept (since \( \ln(\sigma) < 0 \)).

Thus, the correct graph is option **C**.