The timekeeping of an old clock is regulated by a brass pendulum 20.0cm long.

If the clock is accurate at 20 degrees celsius but is in a room at 18 degrees celsius, how long will it be before the clock is in error by one minute? I know the formula for period is 2 X pi X square root (L/g) but how do I relate it to temperature?

Change in temperature affects the length of the pendulum. The amount of change of pendulum made of brass is 0.0000187 m/m k.

First calculate the length of the pendulum in 20 degrees. We know the period of the pendulum is 1s

1 = 2π×sqrt(l/g)
1/(2π) = sqrt(l/g)
(1/(2π))^2 = l/g
g×(1/(2π))^2) = 9.8×(1/(2×3.14))^2) =0.2482369 m.

The change in length for two degrees is 0.0000187*2*0.25 = 0.00000935 m.
Length of pendulum in 18 degree is 0.2482369 - 0.00000935 = 0.24822755.
Period is 2*pi*sqrt(0.24822755/g) = 0.999981167. The amount of time for delay of one minute is:
1 / (1 - 0.999981167) = 53 098.2849 minutes = about 37 days

You need to determine the change in the length. First look up the response of brass to temperature:

http://www.engineeringtoolbox.com/linear…

The expansion coefficient for brass (alpha) is 18.7 10^-6m/m K

dL = L(0) * alpha * dT
dL = 20cm * 18.7 10^-6m/m K * -2K
dL = -0.000748cm
L' = L + dL
L' = 19.999252cm

That's an odd problem in that they gave you the initial length only to 3 significant digits, but you'd need 6 just to see the change in length.