Show that 60.0 L of gasoline originally at 15.0ºC will expand to 61.1 L when it warms to 35.0ºC, how much gasoline has spilled by the time they warm to 35.0ºC ?

To solve this problem, we can use the formula for volume expansion of a substance:

V2 = V1 * (1 + α * ΔT)

where:
V1 = initial volume of the substance (60 L)
V2 = final volume of the substance (61.1 L)
α = coefficient of volume expansion for the substance (assumed to be constant for gasoline)
ΔT = change in temperature (35.0ºC - 15.0ºC = 20ºC)

First, let's isolate α:

α = (V2 / V1 - 1) / ΔT

Now we can substitute the known values into the equation:

α = (61.1 L / 60.0 L - 1) / 20ºC

Simplifying:

α = (1.01833 - 1) / 20ºC
α = 0.01833 / 20ºC
α = 0.0009165 ºC^(-1)

Now we have calculated the coefficient of volume expansion (α) for gasoline.

Next, we can determine the amount of spilled gasoline at 35.0ºC. Since the volume increased from 60.0 L to 61.1 L, the difference represents the amount of spilled gasoline:

Spilled gasoline = V2 - V1
Spilled gasoline = 61.1 L - 60.0 L
Spilled gasoline = 1.1 L

Therefore, by the time the gasoline warms to 35.0ºC, approximately 1.1 liters of gasoline would have spilled.

To find the amount of gasoline spilled when it warms from 15.0ºC to 35.0ºC, we need to calculate the increase in volume of gasoline due to the temperature change.

First, we can determine the initial volume of gasoline at 15.0ºC. Given that the initial volume is 60.0 L.

Next, we can calculate the final volume of gasoline at 35.0ºC. Given that the final volume is 61.1 L.

Now, we can calculate the difference in volume between the initial and final states:

Difference in volume = Final volume - Initial volume
Difference in volume = 61.1 L - 60.0 L
Difference in volume = 1.1 L

Therefore, by the time the gasoline warms to 35.0ºC, 1.1 L of gasoline has spilled.