Q: A copper wire 2 m long is stretched by 1 mm. If the energy stores in the stretched wire is converted into heat, then calculate the rise in temperature of the wire.

(Y= 12.5 × 10^{10} N/m^{2}; ρ = 9 × 10^{3} kg /m^{3} ; s = 385 J/kg – K)

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Sol: $\large m s \Delta T = \frac{1}{2} Stress \times Strain \times Volume $

$\large m s \Delta T = \frac{1}{2} (Y \times Strain) \times Strain \times \frac{m}{\rho} $

$\large \Delta T = \frac{Y}{2 \rho s } \times (\frac{\Delta l}{l})^2$

$\large \Delta T = \frac{12.5 \times 10^{10}}{2 \times 9 \times 10^3 } \times (\frac{10^{-3}}{2})^2$

= 0.0045 ° C

So the rise in temperature of wire is 0.0045 °C.