The Question is from neet-physics thermal-properties-of-matter , and it was created by the author ABDUL JILANI BASHA

Question

A thermocouple using iron and copper produces an emf of $E_1$ when its junctions are kept at $0^\circ C$ and $100^\circ C$.  If the cold junction is kept at $-20^\circ C$ and the emf is $E_2$, what is the temperature of the hot junction?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Thermoelectricity - Seebeck Effect
	**Related Topics**: Temperature Measurement, Thermocouples, Linearity of Thermocouple EMF
	**Prerequisites**: Concept of EMF, Temperature Scales
## SOLUTION PATHWAY
**Given Information**:
	1. EMF $E_1$ when temperatures are $0^\circ C$ and $100^\circ C$.
	2. EMF $E_2$ when cold junction is at $-20^\circ C$.
**Method & Steps**:
	1. **Establish the relationship between EMF and temperature difference:**
		Reasoning: For a small temperature range, the thermo-emf produced by a thermocouple is directly proportional to the temperature difference between the junctions.
		Formula/Rule: $E = \alpha \Delta T$, where $E$ is the emf, $\alpha$ is the Seebeck coefficient, and $\Delta T$ is the temperature difference.
	2. **Apply the relationship to the first case:**
		Working: $E_1 = \alpha (100 - 0) = 100\alpha$
		Key Point: This allows us to express $\alpha$ in terms of $E_1$. $\alpha = \frac{E_1}{100}$
	3. **Apply the relationship to the second case:**
		Working: $E_2 = \alpha (T_h - (-20))$, where $T_h$ is the temperature of the hot junction.
		Substituting $\alpha$ from step 2: $E_2 = \frac{E_1}{100}(T_h + 20)$
	4. **Solve for $T_h$**: 
		Result: $T_h = \frac{100E_2}{E_1} - 20$
		Verification: If $E_2 = E_1$, then $T_h = 100 - 20 = 80^\circ C$.  This makes sense, as a lower cold junction temperature with the same EMF would imply a lower hot junction temperature.
## OPTION ANALYSIS
[A] ✓: $\frac{100E_2}{E_1} - 20$: This is the correct expression derived above.
[B] :x: $\frac{100E_1}{E_2} - 20$: Incorrect. This inverts the ratio of $E_2$ and $E_1$.
[C] :x: $\frac{100E_2}{E_1} + 20$: Incorrect. The sign of 20 should be negative, not positive.
[D] :x: $\frac{E_2}{E_1} + 20$: Incorrect. This is missing the factor of 100 in the numerator.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrect Sign: Pay attention to the sign of the cold junction temperature.
		2. Inverting the ratio of EMFs: Ensure the correct EMF is in the numerator and denominator.
	**Quick Tips**:
		1. Remember the linear relationship between EMF and temperature difference for thermocouples within a limited temperature range.
		2. Always double-check the units and signs in the final expression.

ABDUL JILANI BASHA · Answer

The correct Option is "\frac{100E_2}{E_1} - 20"

ABDUL JILANI BASHA · Answer

\frac{100E_1}{E_2} - 20

ABDUL JILANI BASHA · Answer

\frac{100E_2}{E_1} + 20

ABDUL JILANI BASHA · Answer

\frac{E_2}{E_1} + 20

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