Question

How much does a length of aluminum expand if it is 1.2 meters long at 15 degrees c and heated to 65 degrees c

Answer 1

∆L = aL∆T)
a = 23 × 10^(-6) K^(-1)
L = 1.2 m
∆T = 65°C - 15°C = 50 °C = 323.15 K
∆L = 23 × 10^(-6) × 1.2 × 323.15 m
∆L = 8.91894 × 10^(-3) m