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Thermal Properties Of Matter

Question

When a current is passed in a conductor, 5°C rise in temperature is observed. If the strength of current is made thrice, rise in temperature will be

5 ^oC

20 ^oC

45 ^oC

15 ^oC

Solution

45 ^oC

Joule heating in a wire

H = I² Rt

For given R and t

H ∝ t²

∴ $\frac{H_{1}}{H_{2 I}} = \frac{I_{1}^{2}}{I_{2}^{2}}$ ....(i)

Also H = m s Δ

For given m and s

∴ $\frac{H_{1}}{H_{2}} = \frac{∆ T_{1}}{∆ T_{2}}$ ....(ii)

From equation (i) and (ii)

$\frac{I_{1}^{2}}{I_{2}^{2}} = \frac{∆ T_{1}}{∆ T_{2}}$

Here I₁ = I

ΔT₁ = 5 ^oC

I₂ = 3I

ΔT₂ = ?

∴ ${(\frac{I}{3 I})}^{2} = \frac{5}{∆ T_{2}}$

⇒ ΔT₂ = 45 ^oC