Question

An aluminum alloy rod has a length of 6.3243 cm at 16.00°C and a length of 6.3568 cm at the boiling point of water. (a) What is the length of the rod at the freezing point of water? (b) What is the temperature if the length of the rod is 6.3689 cm?

Answer 1

Answer:

Part a)

$L_o = 6.3181 cm$

Part b)

$T = 131.3 ^oC$

Explanation:

Let the length of the rod at 0 degree Celsius is given as Lo

now we have

$L = L_o( 1 + \alpha \Delta T)$

now we know that

$L_o$ = Length of rod at zero degree C

Part a)

$6.3243 = L_o(1 + \alpha (16 - 0))$

$6.3568 = L_o(1 + \alpha (100 - 0))$

now we have

$\frac{6.3568}{6.3243} = \frac{1 + 100 \alpha}{1 + 16 \alpha}$

$1.005(1 + 16 \alpha) = 1 + 100 \alpha$

$83.918\alpha = 5.138\times 10^{-3}$

$\alpha = 6.12 \times 10^{-5}$

now we have

$L_o = 6.3181 cm$

Part b)

length of the rod is 6.3689 cm

now we have

$L = L_o(1 + \alpha\Delta T)$

$6.3689 = 6.3181(1 + \alpha \Delta T)$

$6.3689 = 6.3181(1 + (6.12 \times 10^{-5})(T - 0))$

$1.008 = 1 + (6.12 \times 10^{-5})T$

$T = 131.3 ^oC$