Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The latent heat of vaporization of liquid nitrogen is about \(200 . \mathrm{kJ} / \mathrm{kg} .\) Suppose you have \(1.00 \mathrm{~kg}\) of liquid nitrogen boiling at \(77.0 \mathrm{~K}\). If you supply heat at a constant rate of \(10.0 \mathrm{~W}\) via an electric heater immersed in the liquid nitrogen, how long will it take to vaporize all of it? What is the time for \(1.00 \mathrm{~kg}\) of liquid helium, whose heat of vaporization is \(20.9 \mathrm{~kJ} / \mathrm{kg}\) ?

Short Answer

Expert verified
Answer: It would take 20,000 seconds to vaporize all of the liquid nitrogen and 2,090 seconds to vaporize all of the liquid helium using a 10 W electric heater.

Step by step solution

01

Convert units

To make calculations, we first need to convert given values into appropriate units. The latent heat of liquid nitrogen is \(200\, \mathrm{kJ/kg}\) which is equal to \((200 \times 10^3) \, \mathrm{J/kg}\). The latent heat of liquid helium is \(20.9\, \mathrm{kJ/kg}\) which is equal to \((20.9 \times 10^3) \, \mathrm{J/kg}\).
02

Calculate heat required for vaporization

We can find the heat required for vaporization using the formula: \(Q = mL\), where \(Q\) is the heat required, \(m\) is the mass, and \(L\) is the latent heat. For liquid nitrogen: \(Q_{\text{N}_2} = (1.0\, \mathrm{kg}) \times (200 \times 10^3 \, \mathrm{J/kg}) = 200 \times 10^3\, \mathrm{J}\). For liquid helium: \(Q_{\text{He}} = (1.0\, \mathrm{kg}) \times (20.9 \times 10^3\, \mathrm{J/kg}) = 20.9 \times 10^3\, \mathrm{J}\).
03

Calculate the time required for vaporization

To find the time required for vaporization, we use the formula: \(t = Q / P\), where \(t\) is the time in seconds, \(Q\) is the heat required for vaporization, and \(P\) is the power provided by the electric heater. For liquid nitrogen: \(t_{\text{N}_2} = \frac{200 \times 10^3\, \mathrm{J}}{10.0\, \mathrm{W}} = 20,000\, \mathrm{s}\). For liquid helium: \(t_{\text{He}} = \frac{20.9 \times 10^3\, \mathrm{J}}{10.0\, \mathrm{W}} = 2,090\, \mathrm{s}\). In conclusion, it would take 20,000 seconds to vaporize all of the liquid nitrogen and 2,090 seconds to vaporize all of the liquid helium using a 10 W electric heater.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Thermodynamics and Latent Heat
In the realm of physics, thermodynamics is the branch that deals with the relationships between heat and other forms of energy. Specifically in the context of latent heat of vaporization, thermodynamics involves the analysis of energy transfer when a substance changes phases from liquid to gas. Latent heat itself is a form of energy released or absorbed by a substance during a phase change without changing its temperature.

The problem involving liquid nitrogen and helium illustrates an important thermodynamic phenomenon where the substances require a certain amount of heat energy, known as the latent heat of vaporization, to transform from a liquid into a vapor. This energy breaks the intermolecular bonds holding the liquid together, allowing molecules to escape into the air as gas. Each substance has a unique latent heat of vaporization value which is crucial in calculating the time taken for the phase change at a given power input from an electric heater.
Heat Transfer Analysis in Vaporization
Heat transfer is a fundamental concept within thermodynamics and engineering that describes the movement of heat energy from one place to another. It plays a crucial role in the process of vaporization, as heat needs to be continuously supplied for a phase change to happen.

From a practical perspective, when heat is supplied at a constant rate via an electric heater, as in the given exercise, it's essential to understand how this energy input affects the speed of vaporization for different substances. The concept of heat transfer efficiency and the ability of a substance to absorb heat (specific heat capacity) are underlying factors determining the rate at which a liquid will be converted into a gas. In solving the exercise, the step-by-step approach breaks down the transfer process into manageable calculations, allowing students to grasp how time, power and energy interact during vaporization.
Phase Change Dynamics
A phase change is a transition of matter from one state to another due to a change in the environmental conditions, primarily temperature and pressure. The provided exercise demonstrates a specific phase change—the vaporization of a liquid into gas. Vaporization requires the liquid to gain enough thermal energy to overcome atmospheric pressure and transform into vapor, a process characterized by the latent heat of vaporization.

While working through similar exercises, students can benefit from a deeper understanding of the phase diagram and the Clausius-Clapeyron relation. These concepts provide insight into how different thermodynamic variables are interconnected and affect the phase transitions of substances like nitrogen and helium. Grasping these fundamentals supports more accurate and intuitive predictions about the behavior of materials under varying thermal conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An air-cooled motorcycle engine loses a significant amount of heat through thermal radiation according to the Stefan-Boltzmann equation. Assume that the ambient temperature is \(T_{0}=27^{\circ} \mathrm{C}(300 \mathrm{~K})\). Suppose the engine generates 15 hp \((11 \mathrm{~kW})\) of power and, due to several deep surface fins, has a surface area of \(A=0.50 \mathrm{~m}^{2}\). A shiny engine has an emissivity \(e=0.050\), whereas an engine that is painted black has \(e=0.95 .\) Determine the equilibrium temperatures for the black engine and the shiny engine. (Assume that radiation is the only mode by which heat is dissipated from the engine.)

How would the rate of heat transfer between a thermal reservoir at a higher temperature and one at a lower temperature differ if the reservoirs were in contact with a 10 -cm-long glass rod instead of a 10 -m-long aluminum rod having an identical cross-sectional area?

Why does tile feel so much colder to your feet after a bath than a bath rug? Why is this effect more striking when your feet are cold?

Which surface should you set a pot on to keep it hotter for a longer time? a) a smooth glass surface b) a smooth steel surface c) a smooth wood surface d) a rough wood surface

When an immersion glass thermometer is used to measure the temperature of a liquid, the temperature reading will be affected by an error due to heat transfer between the liquid and the thermometer. Suppose you want to measure the temperature of \(6.00 \mathrm{~mL}\) of water in a Pyrex glass vial thermally insulated from the environment. The empty vial has a mass of \(5.00 \mathrm{~g}\). The thermometer you use is made of Pyrex glass as well and has a mass of \(15.0 \mathrm{~g}\), of which \(4.00 \mathrm{~g}\) is the mercury inside the thermometer. The thermometer is initially at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) You place the thermometer in the water in the vial and, after a while, you read an equilibrium temperature of \(29.0^{\circ} \mathrm{C} .\) What was the actual temperature of the water in the vial before the temperature was measured? The specific heat capacity of Pyrex glass around room temperature is \(800 . J /(\mathrm{kg} \mathrm{K})\) and that of liquid mercury at room temperature is \(140 . \mathrm{J} /(\mathrm{kg} \mathrm{K})\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free