A lead ball has a volume of 94.3 cm3 at 19.3°C. What is the change in volume when its temperature changes to 34.3°C?
ΔT =34.3-19.3=15º
V=Vₒ(1+3•α•ΔT) =
=94.3•10^-6• (1+3•29•10^-6•15)=
=94.4•10^-6 m³=94.4 cm³.
To find the change in volume of the lead ball when the temperature changes, we can use the coefficient of thermal expansion (α) of lead.
The formula to calculate the change in volume (ΔV) is given by:
ΔV = V * α * ΔT
Where:
V is the initial volume of the lead ball
α is the coefficient of thermal expansion of lead
ΔT is the change in temperature
First, let's determine the initial volume (V) of the lead ball. It is given as 94.3 cm^3.
Next, we need to find the coefficient of thermal expansion (α) of lead. The coefficient of thermal expansion for lead is approximately 0.000029 (1/°C). This means that for every 1°C change in temperature, the volume of lead increases by 0.000029 times its original volume.
Finally, we need to calculate the change in temperature (ΔT). The initial temperature is given as 19.3°C, and the final temperature is given as 34.3°C. Therefore, the change in temperature (ΔT) is 34.3°C - 19.3°C = 15°C.
Now, we can substitute the values into the formula to calculate the change in volume (ΔV):
ΔV = 94.3 cm^3 * 0.000029 (1/°C) * 15°C
Calculating this expression, we find:
ΔV = 0.041 cm^3
Therefore, the change in volume of the lead ball when its temperature changes from 19.3°C to 34.3°C is 0.041 cm^3.