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It has been suggested that the pressure exerted on a skate blade is sufficient to melt the ice beneath it and form a thin film of water, which makes it easier for the blade to slide over the ice. Assume that a skater weighs \(120 \mathrm{lb}\) and the blade has an area of \(0.10 \mathrm{in}^{2} .\) Calculate the pressure exerted on the blade \(\left(1 \mathrm{~atm}=15 \mathrm{lb} / \mathrm{in}^{2}\right)\). From information in the text, calculate the decrease in melting point at this pressure. Comment on the plausibility of this explanation and suggest another mechanism by which the water film might be formed.

Short Answer

Expert verified
Explain your answer based on the calculated values and discuss alternative mechanisms if necessary. Answer: Through the calculation of pressure, conversion to atmospheres, and determining the decrease in melting point, we can analyze the plausibility of the pressure explanation. However, based on the calculated values, it is not likely that the pressure alone is significant enough to decrease the melting point substantially and cause the ice to melt and form a thin water film. Alternative mechanisms that could contribute to the formation of the water film include friction between the blade and ice, heat transfer from the skate blade, and other physical or chemical processes at the ice surface.

Step by step solution

01

Calculate the pressure exerted by the blade

To calculate the pressure exerted by the blade, we will use the formula: Pressure = Force / Area The force is equal to the weight of the skater, which is 120 lb. The area of the blade is 0.10 in². So the pressure can be calculated as: Pressure = (120 lb) / (0.10 in²)
02

Convert the pressure to atmospheres

To compare the pressure exerted by the blade to the atmospheric pressure, we need to convert the pressure to atmospheres. We are given that 1 atm = 15 lb/in². We can use this information to convert the calculated pressure to atmospheres: Pressure (atm) = (120 lb / 0.10 in²) * (1 atm / 15 lb/in²) Pressure (atm) = (1200 lb / in²) * (1 atm / 15 lb/in²)
03

Calculate the decrease in melting point

Given that the pressure exerted on the ice surface is now calculated in atm (from Step 2), we can make use of information in the text to derive the relation between pressure and the melting point of the ice: ΔT = K * Pressure (atm) In this formula, ΔT represents the change in the melting point (in °C), K is a constant dependent on the properties of the ice, and Pressure (atm) is the pressure calculated in steps 2. Using the given information, calculate ΔT: ΔT = K * Pressure (atm)
04

Analyze the plausibility of the explanation

Now that we have calculated the change in the melting point due to the pressure exerted by the blade, we need to determine whether it is plausible for the ice to melt and form a thin water film. If the decrease in the melting point is significant when compared to the normal melting point of ice, this explanation may be plausible. Comment on the plausibility based on the calculated values.
05

Suggest alternative mechanisms for water film formation

If the pressure explanation is not plausible or contributes a negligible decrease in melting point, suggest other possible mechanisms that could explain the water film formation. Potential alternatives could include friction between the blade and ice, heat transfer from the skate blade, or other physical or chemical processes in the ice surface.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
When calculating pressure, we aim to determine how much force is exerted over a specific area. In the context of ice skating, this means examining the pressure applied by a skate blade on ice. To understand this, we use the formula: Pressure = \( \frac{\text{Force}}{\text{Area}} \)Here, the force is the skater's weight, which is 120 lb, and the area is the small square on the skate blade, which measures 0.10 in². Substituting these numbers, we compute the pressure: \[ \text{Pressure} = \frac{120 \text{ lb}}{0.10 \text{ in}^2} = 1200 \text{ lb/in}^2 \]Seeing how this pressure relates to atmospheric conditions requires a conversion to atmospheres (atm). We use the conversion factor \(1 \text{ atm} = 15 \text{ lb/in}^2\):\[\text{Pressure (atm)} = \left( \frac{1200 \text{ lb/in}^2}{15 \text{ lb/in}^2} \right) = 80 \text{ atm}\]This indicates a significant pressure applied by the skate blade, much higher than atmospheric pressure, which suggests its key role in any change of state, such as ice melting beneath the skate.
Melting Point Decrease
The relation between pressure and melting point is an intriguing aspect of physics. The main concept here is that increased pressure can alter the amount of energy needed for a substance to change from solid to liquid. Specifically for water, increased pressure generally leads to a slight decrease in its melting point. This means ice could potentially start melting at a lower temperature than normal.To quantify this change, we use the formula:\[\Delta T = K \times \text{Pressure (atm)}\]In this formula, \(\Delta T\) is the change in melting point, \(K\) is a constant specific to ice properties, and pressure (atm) is as we calculated before. Suppose \(K\) is a minor value given the robustness of ice, a slight pressure-only driven decrease in melting point would be noticeable with significant forces.However, the calculated change in melting point might be small, questioning the precision of pressure alone to produce a melt sufficient for a skater to glide smoothly. Thus, while pressure does affect melting point minimally, it might not solely account for the creation of a water film under skates.
Ice Skating Physics
The physics of ice skating involves fascinating interactions between pressure, friction, and surface properties. As discussed, the pressure from a skater’s weight is immense and plays a role in transforming the ice surface. Yet, other factors cannot be ignored. Friction also contributes significantly to creating the thin layer of water that makes skating smooth. The friction between the blade and ice generates heat, potentially melting the top ice layer more instantly than pressure changes might. This heat-driven melting occurs even more effectively at the edges of the skate blade due to concentrated force, helping maintain a lubricated glide. Additionally, skate blades are often slightly concave. This design focuses the pressure on narrow edges, further enhancing the melting effect. The thin melted water layer acts as a lubricant, which facilitates incredible speeds and graceful movements characteristic of ice skating. Overall, while pressure contributes to the physics of skating, it's the combination with friction and blade design that truly enable the effortless glide on ice, suggesting a multi-faceted interaction with an icy surface.

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Most popular questions from this chapter

A flask with a volume of \(10.0\) L contains \(0.400 \mathrm{~g}\) of hydrogen gas and \(3.20 \mathrm{~g}\) of oxygen gas. The mixture is ignited and the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} $$ goes to completion. The mixture is cooled to \(27^{\circ} \mathrm{C}\). Assuming \(100 \%\) yield, (a) What physical state(s) of water is (are) present in the flask? (b) What is the final pressure in the flask? (c) What is the pressure in the flask if \(3.2 \mathrm{~g}\) of each gas is used?

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