/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 An ice skater blows into cupped ... [FREE SOLUTION] | 91Ó°ÊÓ

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An ice skater blows into cupped hands to warm them, yet at lunch blows across a bowl of soup to cool it. How can this be interpreted thermodynamically?

Short Answer

Expert verified
Blowing into hands transfers heat to them; blowing on soup enhances evaporation and heat transfer from the soup, cooling it.

Step by step solution

01

Understand the Context

The skater uses blowing in two different scenarios: to warm hands and to cool soup. Understanding the surrounding conditions and the nature of the objects involved (hands and soup) is crucial.
02

Analyze Blowing into Cupped Hands

When blowing into cupped hands, the air being blown is warmer than the cold hands. This warmer air warms the hands by transferring heat via convection. The surrounding air is typically much colder than the exhaled breath.
03

Analyze Blowing across Soup

Blowing across hot soup cools it down. The blowing increases the rate of evaporation and enhances convective heat transfer, taking heat away from the soup. Blown air is usually at room temperature, which is cooler than the hot soup.
04

Thermodynamic Interpretation

In both cases, heat transfer principles (convection and evaporation) are applied. For hands, heat transfer is from the warmer air to the colder hands. For soup, heat transfer is from the hot soup to the cooler air and increased evaporation removes heat.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics. It refers to the process of thermal energy moving from a warmer object to a cooler one.
There are three main methods of heat transfer: conduction, convection, and radiation. Blowing air into cupped hands or across hot soup mainly involves convection.
When the ice skater blows into their hands, the warm air from their breath transfers heat to the colder skin. This is an example of heat transfer via convection.
In contrast, blowing over hot soup increases the heat transfer process, cooling the soup down through enhanced convection and evaporation. Therefore, understanding the basics of heat transfer helps explain how blowing air can either warm or cool objects.
Convection
Convection is the process of heat transfer through the movement of fluids or gases. It plays a crucial role in many everyday phenomena.
When the skater blows warm air into cupped hands, this is called forced convection. The warm breath heats the hands, transferring thermal energy from the warmer breath to the cooler skin.
On the other hand, blowing across hot soup also causes convection but results in cooling. By blowing over the surface, air movement picks up heat from the soup, carrying it away and reducing the soup's temperature.
Convective heat transfer is a key concept in understanding many natural and engineered systems, making it essential to grasp thoroughly.
Evaporation
Evaporation is the process where liquid turns into vapor, absorbing heat from the surrounding environment. This process can significantly cool down the liquid.
Blowing across a hot bowl of soup increases the rate of evaporation. As air moves over the soup, it helps more water molecules escape into the air, taking heat away in the process. This is why blowing across soup can cool it.
Evaporation is a fundamental process in many daily activities and industrial applications. It's heavily influenced by factors like temperature, surface area, and air movement.
Understanding evaporation helps explain how heat energy is removed from objects, aiding in practical applications like cooling.

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Most popular questions from this chapter

A gas is contained in a closed rigid tank. An electric resistor in the tank transfers energy to the gas at a constant rate of \(1000 \mathrm{~W}\). Heat transfer between the gas and the surroundings occurs at a rate of \(\dot{Q}=-50 t\), where \(\dot{Q}\) is in watts, and \(t\) is time, in min. (a) Plot the time rate of change of energy of the gas for \(0 \leq t \leq 20 \mathrm{~min}\), in watts. (b) Determine the net change in energy of the gas after 20 min, in kJ. (c) If electricity is valued at \(\$ 0.08\) per \(\mathrm{kW} \cdot \mathrm{h}\), what is the cost of the electrical input to the resistor for \(20 \mathrm{~min}\) of operation?

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