171k views
4 votes
A thin, circular disc is made of lead and has a radius of 0.250 cm at 20.0 °C. Determine the change in the area of the circle if the temperature is increased to 800.0 °C. The coefficient of linear thermal expansion for lead is 29.0 x 10^-6/C°.

2 Answers

3 votes

Final answer:

The change in area of the lead circular disc can be calculated using the formulas for linear thermal expansion. The change in radius is calculated using the coefficient of linear expansion and the change in temperature. The change in area is then calculated using the change in radius and the original radius.

Step-by-step explanation:

To determine the change in area of the circular disc made of lead, we need to calculate the change in its radius using the coefficient of linear thermal expansion. The formula for linear thermal expansion is given by AL = a * L * AT, where AL is the change in length, a is the coefficient of linear expansion, L is the original length, and AT is the change in temperature.

In this case, we are interested in the change in radius, so we can use the formula AR = a * R * AT, where AR is the change in radius and R is the original radius.

Substituting the given values, we have:

AR = (29.0 × 10^-6/°C) * (0.250 cm) * (800.0 °C - 20.0 °C)

AR = 0.00145 cm

The change in area can be calculated using the formula AE = 2π * R * AR. Substituting the values, we have:

AE = 2π * (0.250 cm) * (0.00145 cm)

AE = 2.31 x 10^-3 cm²

User Sandeep Joshi
by
5.9k points
3 votes

Answer:

The change in the area of the circle is
8.88*10^(-7)\ m^2

Step-by-step explanation:

Given that,

Radius = 0.250 cm

Temperature = 20.0°C

Final temperature =800.0°C

Coefficient of linear thermal expansion for lead
\alpha =29.0*10^(-6)\ /\°C

We calculate the change in temperature,


\Delta T=800.0-20.0=780^(\circ)

Now, We calculate the area of the disc


A = \pi r^2

Put the value into the formula


A=3.14*(2.5*10^(-3))^2


A =1.9625*10^(-5)\ m^2

We need to calculate the areal expansion


\Delta A=2\alpha* A*\Delta T


\Delta A=2*29.0*10^(-6)*1.9625*10^(-5)*780


\Delta A=8.88*10^(-7)\ m^2

Hence, The change in the area of the circle is
8.88*10^(-7)\ m^2

User Palm Snow
by
5.7k points