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Intro Page

http://weelookang.blogspot.sg/2014/11/ejss-cube-block-cooling-model.html

image of

copper shiny https://c1.staticflickr.com/1/164/362133253_77585f5429_z.jpg?zz=1

copper dull https://www.colourbox.com/preview/10760507-196481-golden-copper-shiny-abstract-vertical-background.jpg

al shiny http://preview.cutcaster.com/cutcaster-photo-100709683-metal-texture.jpg

al dull http://pixabay.com/p-432524/?no_redirect

fe shiny http://www.burningwell.org/gallery2/d/11247-6/img_0571.jpg

fe dull http://upload.wikimedia.org/wikipedia/commons/1/1d/Old_dirty_dusty_rusty_scratched_metal_iron.jpg

 

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This email address is being protected from spambots. You need JavaScript enabled to view it.; christian wolfgang

http://iwant2study.org/lookangejss/03thermalphysics_11thermalpropertiesofmatter/ejss_model_cooling/cooling_Simulation.xhtml

Briefing Document: 🌡️Block Mass Cooling and Heating Curve Simulation

1. Overview:

This document analyzes a simulation model focusing on heat transfer, specifically the cooling and heating of a block of various materials. The model, developed using Easy JavaScript Simulations (EJS), allows users to explore the principles of thermal physics, particularly Newton's Law of Cooling, and the effects of different materials and surface areas on heat transfer rates. This resource is intended for secondary and junior college level physics education and is accessible on various platforms including computers and mobile devices. The simulation is a valuable tool for visualizing and understanding concepts related to heat conduction, specific heat capacity, and thermal equilibrium.

2. Main Themes and Key Concepts

  • Heat Transfer: The core theme is the transfer of thermal energy between an object and its surroundings. This includes both heating (energy flowing into the object) and cooling (energy flowing out of the object).
  • Newton's Law of Cooling: The simulation is explicitly based on Newton's Law of Cooling, which states that "the rate of thermal energy transfer at the object's surface is proportional to the surface area and to the temperature difference between the object and the surrounding medium." The equation provided is:
  • \frac{\delta Q}{\delta t} = h A( T(t) - T_{background} )
  • Where:
  • \frac{\delta Q}{\delta t} is the rate of heat transfer.
  • h is the heat transfer coefficient.
  • A is the surface area.
  • T(t) is the temperature of the object at time t.
  • T_{background} is the temperature of the surroundings.
  • Specific Heat Capacity: The document defines specific heat capacity as the amount of heat energy required to raise the temperature of 1 kg of a substance by 1 degree Celsius. The formula is:
  • Q = mc ( T_{final} - T_{initial} )
  • Where:
  • Q is the heat energy.
  • m is the mass of the body.
  • c is the specific heat capacity of the material.
  • T_{final} is the final temperature.
  • T_{initial} is the initial temperature.
  • Material Properties: The simulation emphasizes how different materials impact heat transfer. It provides specific heat capacities (c), densities (\rho), and heat transfer coefficients (h) for various materials such as copper, aluminum, and iron, with both shiny and dull surface finishes. Shiny surfaces are assumed to have a higher heat transfer coefficient than dull surfaces: "The Newton's Law of Cooling model assumes h=400 for all shiny and h=200 for dull materials."
  • Surface Area: The model takes into account the surface area of the object, which directly influences heat transfer rates. It makes a calculation of the surface area of a cube assuming that the mass of the object and its density is known. It also introduces a surface area with an increased surface area due to fins:

A = 6 (\frac{m}{\rho})^{\frac{2}{3}} and A_{increased surface area due to fins} = (2)(6) (\frac{m}{\rho})^{\frac{2}{3}}

  • Lumped System Approximation: The model assumes a uniform temperature within the object, which simplifies the calculations. This is valid when "the rate of thermal energy transfer within the object is faster than the rate of thermal energy transfer at the surface."
  • Thermal Equilibrium: The simulation demonstrates that heat transfer ceases when the object and its surroundings reach thermal equilibrium (i.e., the same temperature). As mentioned in the sample lesson notes, "The temperature of the water in both cups remain constant after a while because there is no heat transfer between the watersurrounding and the watersurrounding"

3. Important Facts and Ideas

  • Conductors vs. Insulators: The sample lesson plan within the provided text makes a distinction between good conductors of heat (like steel and copper) and poor conductors of heat (like plastic and wood). The results for the provided experiment in this sample lesson notes show the heat transfers faster with metal than porcelain. "The temperature of the hot water in the metal cup drops faster than the temperature of the hot water in the porcelain cup."
  • Impact of Surface Finish: The simulation and supporting documentation explicitly show that the surface finish (shiny vs. dull) affects the heat transfer coefficient. Shiny surfaces are modeled as having a higher heat transfer coefficient.
  • Mathematical Modeling: The document rigorously derives the differential equation that governs the temperature change in the object, combining Newton's Law of Cooling with the specific heat capacity formula:
  • \frac{ ( T(t) ) }{\delta t} = -\kappa ( T(t) - T_{background} ) where ( \kappa = \frac{h A}{mc } )
  • Adding Heating: The final ODE equation also looks at adding heat, and looks like: \frac{ ( T(t) ) }{\delta t} = -\kappa ( T(t) - T_{background} ) + \frac{heating}{mc}

4. Quotes

  • On Newton's Law of Cooling: "Newton assumed that the rate of thermal energy transfer at the object's surface is proportional to the surface area and to the temperature difference between the object and the surrounding medium."
  • On the Lumped System Assumption: "This lumped system approximation is valid if the rate of thermal energy transfer within the object is faster than the rate of thermal energy transfer at the surface."
  • On Shiny vs. Dull Surfaces: "The Newton's Law of Cooling model assumes h=400 for all shiny and h=200 for dull materials."
  • On Thermal Equilibrium "The temperature of the water in both cups remain constant after a while because there is no heat transfer between the watersurrounding and the watersurrounding."

5. Educational Value:

  • Interactive Simulation: The availability of an interactive simulation allows students to experiment with different parameters (material, surface area, heating) and observe the effects on the cooling/heating curve in real time.
  • Visual Representation: The model provides a visual representation of the heat transfer process, making abstract concepts easier to understand.
  • Hands-On Learning: The use of an EJS model supports active learning, allowing students to explore "what if" scenarios.
  • Real-World Applications: The simulation helps students understand how heat transfer principles are used in real-life situations and design, such as the use of heat sinks and fins to dissipate heat.

6. Limitations:

  • Simplified Model: The model is a simplification of reality. The heat transfer coefficient h is given a simplified value for shiny and dull surfaces, and does not take into account parameters such as the material, the fluid velocity, the fluid viscosity and the condition of the object's surface. The uniform temperature assumption and cubic shape assumption may not hold in all real-world scenarios.
  • Assumed Constant Heat Transfer Coefficient (h): The model assumes that the heat transfer coefficient is independent of the object's temperature, which isn't always true in practice.
  • No Conduction: The model focuses primarily on convective heat transfer and does not directly model thermal conduction within the object itself.

7. Conclusion

The "Block Mass 0.1 kg Cooling and Heating Curve with Different Materials and Surface Area Model" provides a valuable, interactive resource for learning about heat transfer, Newton’s Law of Cooling, and the impact of material properties on thermal behavior. The model, along with its underlying equations, offers a practical approach for understanding complex heat transfer concepts. While it simplifies some aspects of real-world heat transfer, it serves as an excellent tool for developing a fundamental understanding of thermal physics concepts. The inclusion of various materials, surface finishes, and the ability to apply heating make the simulation a powerful educational asset.

 

Thermal Physics Study Guide

Quiz

Instructions: Answer the following questions in 2-3 sentences each.

  1. According to the simulation, how does the material of an object affect its rate of cooling?
  2. What is the key assumption made by the Newton's Law of Cooling model regarding temperature within an object?
  3. What does the heat transfer coefficient represent, and how does it differ between shiny and dull surfaces in the simulation?
  4. How is the surface area of the cube calculated in this simulation, and what happens when fins are added?
  5. What is specific heat capacity, and how does it relate to the amount of energy required to change an object's temperature?
  6. Explain the concept of thermal equilibrium in the context of this simulation.
  7. How does the simulation model the impact of heating, and how does this affect the final ordinary differential equation (ODE)?
  8. Describe how the simulation can be used to explore and compare the thermal properties of different materials.
  9. In the context of the simulation and the experiment described, what does it mean for a material to be a good conductor?
  10. What factors, not explicitly included in the simulation, might influence the actual value of the heat transfer coefficient?

Quiz Answer Key

  1. The material of the object significantly impacts its cooling rate; good conductors of heat, like metals, lose heat faster than poor conductors, like wood or porcelain. Different materials have different specific heat capacities and thermal conductivity, which dictates how quickly they transfer heat.
  2. The Newton's Law of Cooling model assumes that the temperature within the object is uniform at any given time. This means the model is treating the entire object as a single 'lump' with one temperature instead of accounting for different temperatures within the object.
  3. The heat transfer coefficient (h) indicates how effectively heat is transferred between an object and its surroundings; shiny surfaces have a higher coefficient (400 W/(K m²)) than dull surfaces (200 W/(K m²)) in this simulation, representing their greater ability to transfer heat.
  4. The surface area (A) of the cube is calculated using the formula A = 6(m/ρ)^(2/3), assuming a cubic shape. The inclusion of fins doubles this surface area, increasing it by a factor of two, which increases the cooling or heating.
  5. Specific heat capacity (c) is the amount of heat energy (Q) needed to raise the temperature of one kilogram of a substance by one degree Celsius (or Kelvin); a material with a high specific heat capacity requires more energy to achieve the same temperature change than a material with low specific heat capacity.
  6. Thermal equilibrium is reached when there is no more net heat transfer between the object and its surroundings because they have the same temperature. In the simulation, this is the point where the object's temperature becomes constant over time.
  7. The simulation models heating as an additional term in the final ODE that is added to the cooling equation, representing an input of thermal energy. This heating term affects the object's temperature changes, opposing cooling, and is proportional to the heating input and inversely proportional to the product of mc.
  8. The simulation allows users to change parameters such as the material and its surface, observing how the temperature changes over time. This helps users to compare how different materials heat or cool under the same conditions and how these changes are represented on the graph.
  9. In the context of the simulation, a good conductor is a material that quickly transfers heat to or from its surroundings. This means it will cool down much faster when it is hotter than its surroundings.
  10. The heat transfer coefficient is affected by factors such as the fluid velocity, viscosity, surface roughness of the object and fluid, and external flow conditions like turbulent or laminar flow, all of which are not explicitly modeled within the simulation and are simply assigned specific constant values.

Essay Questions

Instructions: Answer the following questions using information from the provided source material.

  1. Discuss the practical implications of Newton's Law of Cooling, using examples from everyday life and considering the limitations of the model.
  2. Compare and contrast how the material properties and surface conditions of an object affect its heat transfer rate, using evidence from the simulation and the given information about the heat transfer coefficient.
  3. Describe the mathematical model that underlies the simulation of thermal processes, explaining the role of each parameter and the assumptions used to simplify the model.
  4. Analyze how the specific heat capacity and the density of a material affect the rate of its temperature change when it's heated or cooled, and discuss how these properties are factored into the simulation.
  5. Explore the educational value of the simulation, explaining how it can aid in understanding heat transfer concepts and the importance of experimental assumptions in scientific modeling.

Glossary of Key Terms

  • Heat Transfer: The movement of thermal energy from a hotter object or region to a colder one.
  • Thermal Energy: The internal energy of an object due to the kinetic energy of its particles; often referred to as 'heat'.
  • Newton's Law of Cooling: A model stating that the rate of heat transfer is proportional to the temperature difference between an object and its surroundings, also dependent on material and surface area.
  • Specific Heat Capacity (c): The amount of heat required to raise the temperature of one kilogram of a substance by one degree Celsius (or Kelvin).
  • Heat Transfer Coefficient (h): A measure of how effectively heat is transferred between a surface and its surrounding fluid; depends on material and surface condition as well as fluid properties.
  • Thermal Conductivity: A measure of a material's ability to conduct heat.
  • Conductor of Heat: A material that allows heat to flow through it easily.
  • Insulator of Heat: A material that resists the flow of heat.
  • Surface Area (A): The total area of the exposed surface of an object.
  • Density (ρ): A measure of the mass of a substance per unit volume.
  • Thermal Equilibrium: A state where there is no net heat transfer between objects or regions because they are at the same temperature.
  • Ordinary Differential Equation (ODE): An equation that expresses the relationship between a function and its derivatives that is used in the simulation's code to track temperature changes over time.
  • Lumped System Approximation: Simplifying assumption that temperature is uniform within an object.
  • Heat (Q): Thermal energy transferred between objects or regions due to a temperature difference, often measured in joules (J).

Sample Learning Goals

1) compare how conductors of heat lose heat or gain heat
2) explain how conductors of heat play an important part in our lives

SLS Lesson by Kong Su Sze

Good conductors of heat

steel, copper

Poor conductors of heat

plastic, wood

The beginning temperature of the hot water in both cups is ◦c. The room temperature of the experiment is ◦c. 
Both the temperature of the hot water in the 2 cups drop over time. The temperature of the hot water in the metal cup drops  than the temperature of the hot water in the porcelain cup.  The hot water in the metal cup heat to the surrounding . The water in the metal cup takes a  time to reach 30 ◦c compared to the water in the porcelain cup.
 
We can conclude that metal is a conductor of heat while porcelain is a conductor of heat.
The temperature of the water in both cups remain constant after a while because there is no heat transfer between the and the .

Answer Key

 

For Teachers

Newton's Law of Cooling


The Newton's Law of Cooling model computes the temperature of an object of mass M as it is heated or cooled by the surrounding medium.

 

Assumption:

The model assumes that the temperature T within the object is uniform. 

Validity:

 This lumped system approximation is valid if the rate of thermal energy transfer within the object is faster than the rate of thermal energy transfer at the surface. 

Convection-cooling "Newton's law of cooling" Model:

Newton assumed that the rate of thermal energy transfer at the object's surface is proportional to the surface area and to the temperature difference between the object and the surrounding medium. 

\( \frac{\delta Q}{\delta t} = h A( T(t) - T_{background} )\)
 
\( Q \) is the thermal energy in joules
\( h \) is the heat transfer coefficient (assumed independent of T here) (\(\frac{W}{m^{2} K}\))
\( A \) is the heat transfer surface area (\( m^{2} \))
\( T \) is the temperature of the object's surface and interior (since these are the same in this approximation)
\( T_{background} \) is the temperature of the surrounding background environment; i.e. the temperature suitably far from the surface is the time-dependent thermal gradient between environment and object.

Definition Specific Heat Capacity:

The specific heat capacity of a material on a per mass basis is
 
\( Q = mc ( T_{final} - T_{initial} ) \)
\( Q \) is heat energy 
\( m \)  is the mass of the body
\( c \) specific heat capacity of a material
\( T_{final} \) is the \(T_{background}\)
\( T_{initial}\) is the \( T(t) \) 
 
 
combing the 2 equations

\( \frac{mc ( T_{background}- T(t) ) }{\delta t} = h A( T(t) - T_{background} )\)

assuming mc is constant'

\( mc \frac{ \delta ( T_{background}- T(t) ) }{\delta t} = h A( T(t) - T_{background} )\)

assuming \(T_{background}\) is a infinite reservoir

\( \frac{ ( T_{background}) }{\delta t} = 0 \)
therefore
\( mc \frac{ ( \delta (- T(t)) ) }{\delta t} = h A( T(t) - T_{background} )\)

negative sign can be taken out of the differential equation.

\( mc \frac{ (\delta T(t) ) }{\delta t} =  -h A( T(t) - T_{background} )\)
 
\( \frac{ ( T(t) ) }{\delta t} =  -\frac{h A}{mc }( T(t) - T_{background} )\)

let \( \kappa  = \frac{h A}{mc } \)

\( \frac{ ( T(t) ) }{\delta t} = -\kappa ( T(t) - T_{background} )\)

If heating is added on,

\( heating = \frac{\delta Q}{\delta t} = mc ( \frac{\delta T}{\delta t}) \)

the final ODE equation looks like

\( \frac{ ( T(t) ) }{\delta t} = -\kappa ( T(t) - T_{background} ) + \frac{heating}{mc}\)

Definition Equation Used:

\( V = \frac{m}{\rho} \)

\( V \) is volume of object
\( \rho \) is density of object


\( A = 6 (\frac{m}{\rho})^{\frac{2}{3}} \)

\( A \) surface area of object

assumption of increased surface are

\( A_{increased surface area due to fins} = (2)(6) (\frac{m}{\rho})^{\frac{2}{3}} \)

Materials added:

copper shiny \( c_{Cu} \) = 385  \( \frac{J}{kg K}\)
\( \rho_{Cu} \) = 8933  \( \frac{kg}{m^{3}}\)
heat transfer coefficient  \(h_{Cu}\) = 400 \( \frac{W}{(K m^{2})} \)

copper dull \( c_{Cu} \) = 385  \( \frac{J}{kg K}\)
\( \rho_{Cu} \) = 8933  \( \frac{kg}{m^{3}}\)
heat transfer coefficient  \(h_{Cu}\) = 200 \( \frac{W}{(K m^{2})} \)

aluminium shiny \( c_{Al} \) = 903  \( \frac{J}{kg K}\)
\( \rho_{Al} \) = 2702  \( \frac{kg}{m^{3}}\)
heat transfer coefficient  \(h_{Al}\) = 400 \( \frac{W}{(K m^{2})} \)

aluminium dull \( c_{Al} \) =  903  \( \frac{J}{kg K}\)
\( \rho_{Al} \) = 2702  \( \frac{kg}{m^{3}}\)
heat transfer coefficient  \(h_{Al}\) = 200 \( \frac{W}{(K m^{2})} \)

iron shiny \( c_{Al} \) = 447  \( \frac{J}{kg K}\)
\( \rho_{Al} \) = 7870  \( \frac{kg}{m^{3}}\)
heat transfer coefficient  \(h_{Al}\) = 400 \( \frac{W}{(K m^{2})} \)

iron dull \( c_{Al} \) =  447  \( \frac{J}{kg K}\)
\( \rho_{Al} \) = 7870  \( \frac{kg}{m^{3}}\)
heat transfer coefficient  \(h_{Al}\) = 200 \( \frac{W}{(K m^{2})} \)
 

Users can select the mass of the object and the material and the model computes the surface area assuming a cubic shape. The model plots the object's temperature as a function of time as the user heats and cools the object. A data-tool button on the temperature graph allows users fit the data to analytic functions. 


Note: A typical (rough) heat transfer coefficient h for still air and iron is 6 W/(K m^2) and 400 W/(K m^2) . The Newton's Law of Cooling model assumes h=400 for all shiny and h=200 for dull materials. The actual value of h depends on many parameters including the material, the fluid velocity, the fluid viscosity and the condition of the object's surface. 

References:


  1. "Measuring the Specific Heat of Metals by Cooling," William Dittrich, The Physics Teacher, (in press).

Credits:

  1. The Newton's Law of Cooling model was created by Wolfgang Christian using the Easy Java Simulations (EJS) version 4.2 authoring and modeling tool.
  2. EJSS Cube Block Cooling Model was created by Wolfgang Christian and recreated by lookang using the Easy Java Simulations (EJS) version 5.1 authoring and modeling tool

Research

[text]

Video

 

https://www.youtube.com/watch?v=8a8nTQIdLOM 

 Version:

  1. http://weelookang.blogspot.sg/2014/11/ejss-cube-block-cooling-model.html 
  2. https://vle.learning.moe.edu.sg/community-gallery/lesson/view/5d907c47-4ae6-47e7-8cce-04199b1cd386

Hands-On Kits

  1. http://www.addest.com/products/category/Science_Kits by Addest Station Cooling Curve Kit Cooling Curve Kit

Other Resources

[text]

FAQ on Heat Transfer and Material Properties

  • How does the material of an object affect its rate of heating or cooling?
  • The material of an object significantly affects how quickly it heats up or cools down due to variations in their specific heat capacity and heat transfer coefficient. Good conductors of heat, such as metals (like copper, aluminum, and iron), tend to gain or lose heat more rapidly than poor conductors, like porcelain, plastic, or wood. This is because materials with higher thermal conductivity allow heat to move more easily through them and out into the surrounding environment. Also, the surface finish of the material (shiny or dull) can affect the heat transfer coefficient, and thus, the rate of heating and cooling. Shiny surfaces typically have a lower heat transfer coefficient than dull surfaces.
  • What is Newton's Law of Cooling, and what assumptions does it make?
  • Newton's Law of Cooling describes the rate at which an object's temperature changes when it is warmer or cooler than its surroundings. It states that the rate of thermal energy transfer at the object's surface is proportional to the surface area and to the temperature difference between the object and its surrounding environment. It assumes that the temperature within the object is uniform (lumped system approximation), which is valid if the rate of thermal energy transfer within the object is much faster than at its surface. It also assumes the heat transfer coefficient is a constant.
  • What is the specific heat capacity of a material, and how is it related to heat transfer?
  • Specific heat capacity (c) is the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1 degree Kelvin (or 1 degree Celsius). Materials with a higher specific heat capacity require more energy to change their temperature, so they heat up and cool down more slowly. Conversely, materials with low specific heat capacities change temperature more readily. The specific heat capacity is a material property and appears in the heat transfer equations.
  • How does the surface area of an object affect its rate of heat transfer?
  • The rate of thermal energy transfer at the object's surface is directly proportional to the surface area (A). This means that an object with a larger surface area will transfer heat at a faster rate than an object with a smaller surface area, assuming all other factors are the same. For a cube, the surface area is calculated as 6 * (m/ρ)^(2/3), where m is the mass and ρ is the density of the material. Increased surface area, such as adding fins, can be modeled as double this surface area.
  • What is the heat transfer coefficient, and how does it vary?
  • The heat transfer coefficient (h) is a measure of how effectively heat is transferred between an object and its surroundings. It represents the proportionality between heat transfer rate and the temperature difference between the object and its surroundings. In the models presented, shiny surfaces are assigned a higher value (400 W/(K m^2)) than dull surfaces (200 W/(K m^2)). However, it's important to note that the actual value of the heat transfer coefficient depends on many factors including the material, the fluid velocity, the fluid viscosity, and the condition of the object's surface and the fluid surrounding it and are not a constant.
  • How is the density of a material related to its surface area and heat transfer?

The density (ρ) of a material is crucial when calculating the surface area of an object, especially when considering shapes like cubes. The surface area is calculated as A = 6*(m/ρ)^(2/3), where m is the mass and ρ is the density of the material. A denser material will result in a smaller surface area for a given mass, thus impacting the rate of heat transfer. Materials with higher density will have less surface area for the same mass.

  • What is the differential equation that describes the temperature change over time for a heated or cooled object in this model?
  • The differential equation describing the temperature change of an object over time, as presented in the model, is:
  • (dT(t)/dt) = -κ(T(t) - T_background) + (heating/mc)
  • where:
  • T(t) is the temperature of the object at time t.
  • T_background is the temperature of the surrounding environment.
  • κ = hA/mc is a constant that incorporates heat transfer coefficient (h), surface area (A), mass (m), and specific heat capacity (c).
  • heating is the rate of heat energy added to the object, if any. This equation incorporates both the cooling or heating due to heat transfer with the background and any added heating.
  • How can simulations, such as the one mentioned, aid in understanding these concepts?
  • Simulations like the Easy JavaScript Simulations (EJS) model provide an interactive way to explore heat transfer concepts. Users can select different materials, adjust parameters (mass, surface area), and see the dynamic temperature changes as the object heats or cools. They allow users to visualize the effects of material properties and surface conditions on heat transfer, which provides a more intuitive learning experience than static equations or conceptual explanations. The simulations can be fitted to analytic curves, deepening a student's understanding of heat transfer concepts.
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