What is the relationship between surface area and heat loss

thermodynamics - Surface Area: Volume and its relation with heat - Physics Stack Exchange

All else equal, heat loss from an object to its surroundings, by convection or radiation, is directly proportional to the objects surface area. Generally, if you. The aim of this experiment is to find out the relationship between surface area and rate How Surface Area to Volume Ratio Affects the Rate of Heat Loss Essay. However, the surface area of the Plasticine is only one of the variables, to I predict the star of Plasticine will heat up and cool down (transfer heat) read more. An investigation into the relationship between heat loss and surface area to.

Such Styrofoam products are made by blowing an inert gas at high pressure into the polystyrene before being injected into the mold. The gas causes the polystyrene to expand, leaving air filled pockets that contribute to the insulating ability of the finished product. Styrofoam is used in coolers, pop can insulators, thermos jugs, and even foam boards for household insulation. Another solid insulator is cellulose. Cellulose insulation is used to insulate attics and walls in homes.

It insulates homes from heat loss as well as sound penetration. It is often blown into attics as loose fill cellulose insulation. It is also applied as fiberglass batts long sheets of paper backed insulation to fill the spacing between 2x4 studs of the exterior and sometimes interior walls of homes. Area Another variable that affects the rate of conductive heat transfer is the area through which heat is being transferred.

For instance, heat transfer through windows of homes is dependent upon the size of the window. More heat will be lost from a home through a larger window than through a smaller window of the same composition and thickness. More heat will be lost from a home through a larger roof than through a smaller roof with the same insulation characteristics.

Each individual particle on the surface of an object is involved in the heat conduction process. An object with a wider area has more surface particles working to conduct heat.

Lecture 06: 1D Steady State Heat Conduction In Plane Wall With Generation of Thermal Energy

As such, the rate of heat transfer is directly proportional to the surface area through which the heat is being conducted. Thickness or Distance A final variable that affects the rate of conductive heat transfer is the distance that the heat must be conducted. Heat escaping through a Styrofoam cup will escape more rapidly through a thin-walled cup than through a thick-walled cup. The rate of heat transfer is inversely proportional to the thickness of the cup. A similar statement can be made for heat being conducted through a layer of cellulose insulation in the wall of a home.

The thicker that the insulation is, the lower the rate of heat transfer. Those of us who live in colder winter climates know this principle quite well.

We are told to dress in layers before going outside. This increases the thickness of the materials through which heat is transferred, as well as trapping pockets of air with high insulation ability between the individual layers. A Mathematical Equation So far we have learned of four variables that affect the rate of heat transfer between two locations.

Investigating the Relationship Between Heat Loss and Surface Area

The variables are the temperature difference between the two locations, the material present between the two locations, the area through which the heat will be transferred, and the distance it must be transferred. As is often the case in physics, the mathematical relationship between these variables and the rate of heat transfer can be expressed in the form of an equation.

Let's consider the transfer of heat through a glass window from the inside of a home with a temperature of T1 to the outside of a home with a temperature of T2. The window has a surface area A and a thickness d.

This equation is applicable to any situation in which heat is transferred in the same direction across a flat rectangular wall. It applies to conduction through windows, flat walls, slopes roofs without any curvatureetc. A slightly different equation applies to conduction through curved walls such as the walls of cans, cups, glasses and pipes. We will not discuss that equation here.

Example Problem To illustrate the use of the above equation, let's calculate the rate of heat transfer on a cold day through a rectangular window that is 1. To solve this problem, we will need to know the surface area of the window. Therefore I deduced that if I wanted to have accurate and reliable results I would have to change to a much larger container i.

Planning After taking into consideration what happened in my preliminary work I decided to fill: A cm3 beaker with cm3 of water.

A cm3 beaker with cm2 of water. And a 40cm3 beaker with 40cm3 of water. I chose these certain volumes for two reasons. Firstly that they are far apart and so will not be similar with each other and have different gradients.

• An investigation into the relationship between heat loss and surface area to volume ratio
• Rates of Heat Transfer

I will have a thermometer clamped into position half way into the beaker. I will do this because, as I learned from my preliminary work, there are localised heat spots. So if the thermometer is clamped into a secure position these localised heat spots will not make the temperature shown on the thermometer oscillate, as the thermometer will not roll in and out of these hot spots. The safety for the heating procedure is simple.

How Surface Area affects Heat Transfer.

But we must still wear safety goggles and an apron to ensure our eyes remain out of harms way. I will repeat each experiment 3 times, then average these results and so plot them onto a graph. This will make the data more reliable and the chance of an anomalous result occurring will be significantly reduced.

The apparatus will be set up as shown below.

I will have to work out the surface area and volume in order to find the surface area to volume ratio. Below is a diagram of a beaker and therefore I will prove why the formula works. The first variable is the volume of the water; this is easily controlled by simply measuring out how much water is being used. The localised heat spots are variables as well. To ensure a fair reduction of these heat spots I simply stirred the beaker with the thermometer while I heated.

During cooling these localised heat spots will be controlled be keeping the thermometer in one place. Thus ensuring no accidental contact with the spots which were spread throughout the beaker.

Rates of Heat Transfer

The temperature of the surrounding air is another variable. I will control this by performing the experiment in one day. This is difficult but must be done. The reason for this is that no day has quite the same temperature and so will affect the cooling rate of the water. Basically this happens because if for example the day is cooler the particles will have less energy than a day which is warmer. Excessive boiling is another variable. What I mean by this is simply that you will lose a relatively considerable amount of water if you boil the water for a long period of time, note only this but during cooling some evaporation will also occur.