A 3.00-liter aluminum cylinder at 5.00°C is filled to the brim with gasoline at the same temperature. If the aluminum and gasoline are warmed to 52.0°C, how much gasoline spills out? [Hint: Be sure to account for the expansion of the container. Also, ignore the possibility of evaporation, and assume the volume coefficients are good to three digits. where the average coefficient of linear expansion of aluminum is 24x10^-6 degrees C^-1 and the average coefficient of volume expansion of gasoline is 9.6x10^-4 degrees C^-1
Well, it sounds like we have quite the hot topic here! Let's see if we can solve this conundrum.
First, we need to determine the change in temperature for both the aluminum cylinder and the gasoline. From 5.00°C to 52.0°C, that's a rise of 47.0°C.
Now, we need to calculate the change in volume for the aluminum cylinder. The coefficient of linear expansion for aluminum is 24x10^-6 degrees C^-1. Multiplying this by the change in temperature in degrees Celsius (47.0°C), we get:
24x10^-6 x 47.0 = 1.128x10^-3.
Since the aluminum cylinder is in the shape of a cylinder, we need to consider the change in volume due to the change in length. The original length is not given, so let's assume it doesn't change. In that case, the change in volume is simply the change in cross-sectional area multiplied by the change in length. Since the length does not change, the change in volume is solely due to the change in cross-sectional area.
The initial cross-sectional area can be calculated using the formula for the area of a circle: A = πr^2, where r is the radius of the cylinder. Since it's filled to the brim, the radius is the same as the original length.
Now, let's find the change in volume for the gasoline. The average coefficient of volume expansion of gasoline is 9.6x10^-4 degrees C^-1. Multiplying this by the change in temperature in degrees Celsius (47.0°C), we get 4.512x10^-2.
So, to find out how much gasoline spills out, we need to subtract the change in volume of the aluminum from the change in volume of the gasoline:
4.512x10^-2 - 1.128x10^-3 = 4.3992x10^-2.
Therefore, approximately 4.3992x10^-2 liters of gasoline would spill out. Just remember, when it comes to spilled gasoline, it's best to handle it with care and avoid any unwanted combustion or slippery situations! Stay safe, my friend!
To solve this problem, we need to consider the expansion of both the aluminum cylinder and the gasoline as they are warmed.
First, let's calculate the change in volume of the aluminum cylinder when it is heated from 5.00°C to 52.0°C.
The change in volume of the aluminum cylinder can be calculated using the formula:
ΔV = V₀ * β * ΔT
Where:
ΔV is the change in volume
V₀ is the initial volume of the aluminum cylinder (3.00 liters)
β is the coefficient of linear expansion of aluminum (24 x 10^-6 degrees C^-1)
ΔT is the change in temperature (52.0°C - 5.00°C)
Calculating the change in volume of the aluminum cylinder:
ΔV_aluminum = V₀ * β * ΔT
ΔV_aluminum = 3.00 L * (24 x 10^-6 1/°C) * (52.0°C - 5.00°C)
ΔV_aluminum ≈ 0.0034 L
Next, let's calculate the change in volume of the gasoline when it is heated from 5.00°C to 52.0°C.
The change in volume of the gasoline can be calculated using the formula:
ΔV = V₀ * β * ΔT
Where:
ΔV is the change in volume
V₀ is the initial volume of the gasoline, which is the same as the volume of the aluminum cylinder (3.00 liters)
β is the coefficient of volume expansion of gasoline (9.6 x 10^-4 degrees C^-1)
ΔT is the change in temperature (52.0°C - 5.00°C)
Calculating the change in volume of the gasoline:
ΔV_gasoline = V₀ * β * ΔT
ΔV_gasoline = 3.00 L * (9.6 x 10^-4 1/°C) * (52.0°C - 5.00°C)
ΔV_gasoline ≈ 0.1714 L
Since the aluminum cylinder expands by 0.0034 liters and the gasoline expands by 0.1714 liters, the total change in volume is the sum of these two values:
Total change in volume = ΔV_aluminum + ΔV_gasoline
Total change in volume ≈ 0.0034 L + 0.1714 L
Total change in volume ≈ 0.1748 L
Therefore, approximately 0.1748 liters of gasoline spills out when the aluminum cylinder and gasoline are warmed from 5.00°C to 52.0°C.