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QPages 218-219: Q14: HC Verma: Concepts of Physics - Volume 2 Thermal and Chemical Effects of Electric Current Pages 218-219: Q14 Solution: Find the thermo-emf developed in a copper-silver thermocouple…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often fail to correctly subtract the Seebeck coefficients of the two metals relative to the reference material (lead), leading to a zero or incorrect $a$ value. Always calculate the net coefficient as the difference between the hot junction metal and the cold junction metal ($a_{net} = a_{hot} - a_{cold}$) to ensure the correct sign and magnitude of the thermo-emf.

Find the thermo-emf developed in a copper-silver thermocouple when the junctions are kept at 0�C and 40�C. Use the data: For Copper with lead, a=2.76 μV/∘Ca = 2.76 \ \mu\text{V}/^\circ\text{C}, b=0.012 μV/∘C)2b = 0.012 \ \mu\text{V}/^\circ\text{C})^2. For Silver with lead, a=2.50 μV/∘Ca = 2.50 \ \mu\text{V}/^\circ\text{C}, b=0.012 μV/∘C)2b = 0.012 \ \mu\text{V}/^\circ\text{C})^2.

Step-by-Step Verified Solution

**Answer:** Thermo-emf = 1.04×10−51.04 \times 10^{-5} V. **Step-by-Step Explanation:** aCu,Ag=aCu,Pb−aAg,Pb=2.76−2.50=0.26 μV/�Ca_{Cu,Ag} = a_{Cu,Pb} - a_{Ag,Pb} = 2.76 - 2.50 = 0.26\ \mu\text{V}/�\text{C}. bCu,Ag=bCu,Pb−bAg,Pb=0.012−0.012=0b_{Cu,Ag} = b_{Cu,Pb} - b_{Ag,Pb} = 0.012 - 0.012 = 0. θ=40�C\theta = 40�\text{C}. E=aθ=0.26×40=10.4 μV=1.04×10−5E = a\theta = 0.26 \times 40 = 10.4\ \mu\text{V} = 1.04 \times 10^{-5} V .

All Questions in this Chapter

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