Intro Page

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

### Translations

Code Language Translator Run

### Software Requirements

SoftwareRequirements

 Android iOS Windows MacOS with best with Chrome Chrome Chrome Chrome support full-screen? Yes. Chrome/Opera No. Firefox/ Samsung Internet Not yet Yes Yes cannot work on some mobile browser that don't understand JavaScript such as..... cannot work on Internet Explorer 9 and below

### Credits

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

### end faq

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

### 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

### Shiny or Dull bug fixed

Title: Block Mass 0.1 kg Cooling and Heating Curve with Different Materials and Surface Area Model

## Introduction:

In this blog post, we will explore the cooling and heating curves of a block with a mass of 0.1 kg, focusing on the influence of different materials and surface area on the temperature changes. Understanding these factors is crucial for various applications, from designing efficient cooling systems to optimizing heating processes. So, let's dive into the fascinating world of thermal dynamics!

## The Basics of Cooling and Heating Curves:

Before we delve into the specifics, let's briefly explain what cooling and heating curves represent. A cooling curve illustrates how the temperature of an object changes over time as it loses heat to its surroundings. Conversely, a heating curve shows the temperature changes as the object gains heat from its surroundings. Both curves typically display a gradual decrease or increase in temperature, eventually reaching a state of equilibrium.

## Exploring Shiny or Dull:

When it comes to cooling down, the characteristics of different surface types, such as shiny and dull, play a significant role. Shiny surfaces, with their smooth and reflective nature, have the ability to reflect a considerable amount of incoming radiation, including heat. This reflective property allows shiny surfaces to repel heat energy, resulting in slower temperature decreases.

On the other hand, dull surfaces, which have a rough and non-reflective texture, absorb more radiation. As a result, dull surfaces retain heat energy and cool down faster compared to shiny surfaces. The contrasting behaviors of shiny and dull surfaces emphasize the importance of surface properties in the cooling process, highlighting how their characteristics can impact the rate at which an object or material cools down.

When exploring different surface types, such as shiny and dull, we encounter distinct characteristics that influence their interaction with heat and light. Shiny surfaces, characterized by a smooth and reflective texture, have the ability to reflect a significant portion of incoming radiation, including heat. As a result, shiny surfaces tend to absorb less heat energy, leading to slower temperature increases. Conversely, dull surfaces, with their rough and non-reflective texture, have lower reflectivity and greater capacity to absorb radiation. This absorption promotes the retention of heat energy, causing dull surfaces to heat up faster compared to their shiny counterparts. The reflective properties of shiny surfaces and the absorptive properties of dull surfaces highlight the importance of surface characteristics when considering heat transfer and temperature changes.

## Exploring Different Materials:

The choice of material significantly affects the cooling and heating behavior of an object.

When both blocks are exposed to a colder environment, such as a room at a lower temperature, the metal block will cool down faster than the wooden block. Metals generally have higher thermal conductivity, allowing heat to transfer more rapidly from the block's surface into the surrounding air. Wood, on the other hand, has lower thermal conductivity, resulting in a slower cooling process.

Conversely, when both blocks are subjected to a hotter environment, such as a heat source, the metal block will heat up faster than the wooden block. Metals again exhibit higher thermal conductivity, enabling them to absorb and distribute heat more efficiently.

## Examining Surface Area:

In addition to material properties, the surface area of an object also plays a crucial role in its cooling and heating rates. To observe this effect, let's consider two blocks made of the same material, both with a mass of 0.1 kg, but with different surface areas.

If we compare a larger block with a larger surface area to a smaller block with a smaller surface area, both exposed to the same cooling or heating conditions, the larger block will experience faster temperature changes. This is because the larger surface area provides more space for heat exchange with the surroundings, promoting quicker heat transfer and resulting in a more rapid cooling or heating process.

Conclusion:

Understanding the cooling and heating curves of objects is essential for numerous real-world applications. In this blog post, we explored how different materials and surface areas can influence the temperature changes of a 0.1 kg block. Metals generally exhibit higher thermal conductivity, leading to faster heat transfer and thus quicker cooling and heating. Additionally, objects with larger surface areas experience more rapid temperature changes due to increased heat exchange with the surroundings.

By considering these factors, engineers and scientists can design more efficient cooling and heating systems, optimize thermal processes, and make informed decisions in various industries. So, next time you encounter a cooling or heating challenge, remember the influence of materials and surface area on the fascinating world of thermal dynamics!

### 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 .

## 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 $$\kappa = \frac{h A}{mc }$$

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

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

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

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

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

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