Calorimeter Homework: Clock Period & Temperature Change

[mimi83]

Homework Statement

A clock is designed with a simple pendulum, consisting of a mass at the end of an aluminum rod, whose period is supposed to be equal to one second. The clock is designed to keep accurate time at 20.0°C, but is operated at a constant temperature of −80.0°C. (a) Will the clock run fast or slow? Explain. (b) How many periods will the pendulum go through before the clock is off by 1.00 s? (The coefficient of linear expansion of aluminum is 2.31×10−5 1/C°.)

Homework Equations

Q = cmΔT
Qw + Qo = 0
ΔL = LiαΔT

The Attempt at a Solution

i tried several time using those equation above, but it didnot give the right answer, please help or give me hints.thank

 

[LowlyPion]

Aren't you wanting to know the length of the pendulum?
http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Rule_of_thumb_for_pendulum_length

 

[nlightner]

you

I would like to first clarify that the equations you have provided are related to calorimetry, which deals with the measurement of heat and temperature changes. However, in this scenario, we are dealing with a clock and its accuracy, which is not directly related to heat and temperature changes.

To solve this problem, we need to consider the effects of temperature on the period of a pendulum. The period of a pendulum is directly proportional to the square root of its length and inversely proportional to the square root of the acceleration due to gravity. However, temperature can also affect the length of the pendulum due to thermal expansion.

(a) In this scenario, the clock is operated at a temperature much lower than its designed temperature. As a result, the aluminum rod will contract due to thermal contraction, causing the pendulum to become shorter. This will result in a shorter period of oscillation, meaning the clock will run faster at -80.0°C compared to 20.0°C.

(b) To calculate the number of periods that the pendulum will go through before the clock is off by 1.00 s, we need to use the equation:

ΔT = (ΔL/Li) * (1/α)

Where ΔT is the change in temperature, ΔL is the change in length, Li is the initial length of the pendulum, and α is the coefficient of thermal expansion of aluminum.

We know that the clock is off by 1.00 s, which means the period of the pendulum has changed by 1.00 s. We also know that the coefficient of thermal expansion of aluminum is 2.31×10−5 1/C°. Therefore, we can rearrange the equation to solve for ΔL:

ΔL = (ΔT * α * Li)

Substituting the values, we get:

ΔL = (1.00 s * 2.31×10−5 1/C° * 1 m) = 2.31×10−5 m

This means that the pendulum has shortened by 2.31×10−5 m at -80.0°C. Now, we can use the equation for the period of a pendulum to calculate the number of periods:

T = 2π * √(L/g)

Where T is the period, L is the length of the pendulum, and g is