At room temperature (27° C), the resistance of a heating element is 50 Ω. The temperature coefficient of the material is 2.4 × 10⁻⁴ ° C⁻¹. The temperature of the element, when its resistance is 62 Ω, is __________ ° C.

Solution

Physics sir · 3 steps

Take 27 °C as your reference, find how much the temperature rose, then add 27 back.

1. Linear resistance rule. For a metal, resistance grows linearly with temperature. Use the known value at room temperature as the reference: R₁ = 50 Ω at T₁ = 27°C.

   R₂ = R₁[1 + α (T₂ - T₁)]

2. Find the rise. 62/50 = 1.24, so α (T₂ - T₁) = 0.24. Then T₂ - T₁ = 0.24/(2.4 × 10⁻⁴) = 1000°C.
3. So. T₂ = 27 + 1000 = 1027°C, so the answer is 1027.

The common slip: 1000. Stopping at the temperature rise gives 1000, not 1027. The 27 °C starting temperature still has to be added.

The idea to keep. R = R₁[1 + α Δ T] gives you the temperature change, not the temperature. Always add the starting temperature at the end.

Answer from the NTA final answer key. Solution by PrepWiser.