- #1
Petar Mali
- 290
- 0
[tex]I_A=BT^2e^{-\frac{e\varphi}{k_BT}}[/tex]
We measure [tex]T[/tex] and [tex]I_A[/tex] and from that get work. If [tex]T_0[/tex] is first temperature in which we measured than
we get
[tex]\frac{I_A}{I_{A0}}=\frac{T^2}{T^2_0}e^{-\frac{e\varphi}{k_BT}(\frac{1}{T}-\frac{1}{T_0})}[/tex]
And from that we get
[tex]\varphi=-\frac{k_B}{e(\frac{1}{T}-\frac{1}{T_0})}ln\frac{I_AT^2_0}{I_{A0}T^2}[/tex]
and from that they calculate tangent like
[tex]\varphi=-\frac{k_B}{e}\frac{\Delta ln\frac{I_AT^2_0}{I_{A0}T^2}}{\Delta \frac{1}{T}}}[/tex]
What happened with [tex]T_0[/tex]?
We measure [tex]T[/tex] and [tex]I_A[/tex] and from that get work. If [tex]T_0[/tex] is first temperature in which we measured than
we get
[tex]\frac{I_A}{I_{A0}}=\frac{T^2}{T^2_0}e^{-\frac{e\varphi}{k_BT}(\frac{1}{T}-\frac{1}{T_0})}[/tex]
And from that we get
[tex]\varphi=-\frac{k_B}{e(\frac{1}{T}-\frac{1}{T_0})}ln\frac{I_AT^2_0}{I_{A0}T^2}[/tex]
and from that they calculate tangent like
[tex]\varphi=-\frac{k_B}{e}\frac{\Delta ln\frac{I_AT^2_0}{I_{A0}T^2}}{\Delta \frac{1}{T}}}[/tex]
What happened with [tex]T_0[/tex]?