Arun
Last Activity: 7 Years ago
Recall that, by linear expansion:
∆L = aL₁∆T.
Solving this for ∆T gives:
∆T = ∆L/(aL₁)
Since the want the lenses to be 2.21 cm and they are initially 2.20 cm, we have:
∆L = 2.21 - 2.20 = 0.01 cm.
Plugging this into our equation gives:
∆T = ∆L/(aL₁)
= (0.01 cm)/[(1.30 x 10^-4° C^-1)(2.20 cm)]
= 34.9°C.
Therefore, the temperature desired is 20.0 + 34.9 = 54.9°C.
I hope this helps!