Verified Solution • 2025–26 Session moderate

QPages 218-219: Q13: HC Verma: Concepts of Physics - Volume 2 Thermal and Chemical Effects of Electric Current Pages 218-219: Q13 Solution: The temperatures of the junctions of a bismuth-silver…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often insert the hot‑junction temperature (0.001 °C) directly into $E=a\theta+\frac12b\theta^2$ instead of using the temperature *difference* $\Delta T=Th-Tc$; the formula requires $\theta=\Delta T$, so compute $\theta=0.001-0=0.001\,$°C before applying the coefficients.

The temperatures of the junctions of a bismuth-silver thermocouple are maintained at 0�C and 0.001�C. Find the thermo-emf Seebeck emf) developed. For bismuth-silver, a=−46×10−6 V∘C−1a = -46 \times 10^{-6} \text{ V}^\circ\text{C}^{-1} and b=−0.48×10−6 V∘C−2b = -0.48 \times 10^{-6} \text{ V}^\circ\text{C}^{-2}.

Step-by-Step Verified Solution

**Answer:** Thermo-emf = −4.6×10−8-4.6 \times 10^{-8} V. **Step-by-Step Explanation:** a=−46×10−6a = -46 \times 10^{-6} V/�C, b=−0.48×10−6b = -0.48 \times 10^{-6} V/�C_, θ=0.001�C\theta = 0.001�\text{C}.
E=aθ+12bθ2=(−46×10−6)(0.001)+12(−0.48×10−6)(0.001)2E = a\theta + \frac{1}{2}b\theta^2 = (-46 \times 10^{-6})(0.001) + \frac{1}{2}(-0.48 \times 10^{-6})(0.001)^2
E=−4.6×10−8 VE = -4.6 \times 10^{-8}\ \text{V}

All Questions in this Chapter

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