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Chapter 17: Problem 55

On a cool morning, with the temperature at \(15.0^{\circ} \mathrm{C}\), a painter fills a 5.00 -gal aluminum container to the brim with turpentine. When the temperature reaches \(27.0^{\circ} \mathrm{C}\), how much fluid spills out of the container? The volume expansion coefficient for this brand of turpentine is \(9.00 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

Short Answer

Expert verified
Answer: Approximately 0.20416 liters of turpentine spill out of the container.

Step by step solution

01

Convert gallons to liters

First, we need to convert the container's volume from gallons to liters, as we will be using the SI unit for volume in our calculations. The conversion factor is 1 gallon ≈ 3.78541 liters. So, the initial volume of the turpentine in liters is: \(V_i = 5.00 \:\text{gal} \times 3.78541 \:\frac{\text{L}}{\text{gal}} = 18.92705 \:\text{L}\)
02

Calculate the temperature change

Next, we need to calculate the change in temperature, which is the difference between the final and initial temperatures: \(\Delta T = T_f - T_i = 27.0^{\circ} \mathrm{C} - 15.0^{\circ} \mathrm{C} = 12.0^{\circ} \mathrm{C}\)
03

Calculate the volume expansion

Now, we can calculate the volume expansion of the turpentine using the volume expansion coefficient, the initial volume, and the change in temperature: \(\Delta V = \beta \times V_i \times \Delta T = (9.00 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}) \times (18.92705 \:\text{L}) \times (12.0^{\circ} \mathrm{C})\) \(\Delta V \approx 0.20416 \:\text{L}\) So, the turpentine expands by approximately 0.20416 liters when the temperature increases.
04

Calculate the amount of fluid spilled

Finally, since the container was initially filled to the brim, the amount of turpentine that spills out of the container is equal to the volume expansion: Spilled turpentine ≈ \(\Delta V \approx 0.20416 \:\text{L}\) The amount of turpentine that spills out of the container when the temperature increases from \(15.0^{\circ} \mathrm{C}\) to \(27.0^{\circ} \mathrm{C}\) is approximately 0.20416 liters.

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Most popular questions from this chapter

The solar corona has a temperature of about \(1 \cdot 10^{6} \mathrm{~K}\). However, a spaceship flying in the corona will not be burned up. Why is this?

Two solid objects, \(A\) and \(B\), are in contact. In which case will thermal energy transfer from \(\mathrm{A}\) to \(\mathrm{B} ?\) a) \(\mathrm{A}\) is at \(20{ }^{\circ} \mathrm{C},\) and \(\mathrm{B}\) is at \(27{ }^{\circ} \mathrm{C}\) b) \(A\) is at \(15^{\circ} \mathrm{C},\) and \(\mathrm{B}\) is at \(15^{\circ} \mathrm{C}\). c) \(\mathrm{A}\) is at \(0{ }^{\circ} \mathrm{C},\) and \(\mathrm{B}\) is at \(-10{ }^{\circ} \mathrm{C}\).

A piece of dry ice (solid carbon dioxide) sitting in a classroom has a temperature of approximately \(-79^{\circ} \mathrm{C}\) a) What is this temperature in kelvins? b) What is this temperature in degrees Fahrenheit?

At room temperature, an iron horseshoe, when dunked into a cylindrical tank of water (radius of \(10.0 \mathrm{~cm})\) causes the water level to rise \(0.25 \mathrm{~cm}\) above the level without the horseshoe in the tank. When heated in the blacksmith's stove from room temperature to a temperature of \(7.00 \cdot 10^{2} \mathrm{~K}\) worked into its final shape, and then dunked back into the water, how much does the water level rise above the "no horseshoe" level (ignore any water that evaporates as the horseshoe enters the water)? Note: The linear expansion coefficient for iron is roughly that of steel: \(11 \cdot 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\).

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