Calculate the increase in the length of an aluminum bar 2 m long when heated from 0°C to 20°C, if the thermal coefficient of linear expansion of aluminum is $2.5 \times 〖10〗^{-5} K^{-1}$. (0.1cm)
Difficulty: Easy
Solution: Original length of rod = $L_{(0=)}$ 2m
Initial temperature = $T_{(0=)}$ 0°C = 0 + 273 = 273 K
Final temperature = T = 20°C = 20 + 273 = 293 K
Change in temperature = $\triangle T$ = $T - T_{0}$ = 293 – 273 = 20 K
Coefficient of linear expansion of aluminum = α = $2.5 \times 10^{-5} K^{-1}$
Increase in volume $\triangle L$ = ?
$ \triangle L = α L_0 \triangle$ T
$\triangle L = 2.5 \times 10^{-5} \times20$
$\triangle L = 100 \times 10^{-5}$
$\triangle L = 0.001 m = 0.001\times$ 100 = 0.1cm
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