Under most circumstances the dominant mode of cooling will be convection, and the rate of heat loss per unit area of skin is given by Newton's. Apr 9, All else equal, heat loss from an object to its surroundings, by convection or radiation, is directly proportional to the objects surface area. Where h is the heat transfer coefficient, A is the surface area, and T_object-T_fluid is the driving temperature difference between the. The relationship between the size of an organism or structure and its surface area . Size - Rate of heat loss is dependent on the surface area of the organism.

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. 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. 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.

**2.1.6 Surface Area to Volume Ratio**

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.

### Rates of Heat Transfer

This happens because as the molecules in the water get enough energy they will be able to escape into the surrounding atmosphere. 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.

## Rates of Heat Transfer

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.

We will also need to give attention to the unit on thickness d.

It is given in units of cm; we will need to convert to units of meters in order for the units to be consistent with that of k and A. The thermal conductivity of glass is about 0.

Glass windows are constructed as double and triple pane windows with a low pressure inert gas layer between the panes. Furthermore, coatings are placed on the windows to improve efficiency. The result is that there are a series of substances through which heat must consecutively pass in order to be transferred out of or into the house. Like electrical resistors placed in seriesa series of thermal insulators has an additive effect on the overall resistance offered to the flow of heat.

The accumulative effect of the various layers of materials in a window leads to an overall conductivity that is much less than a single pane of uncoated glass.

Lesson 1 of this Thermal Physics chapter has focused on the meaning of temperature and heat. Emphasis has been given to the development of a particle model of materials that is capable of explaining the macroscopic observations. Efforts have been made to develop solid conceptual understandings of the topic in the absence of mathematical formulas.

This solid conceptual understanding will serve you well as you approach Lesson 2. The chapter will turn slightly more mathematical as we investigate the question: Lesson 2 will pertain to the science of calorimetry. Check Your Understanding 1. Predict the effect of the following variations upon the rate at which heat is transferred through a rectangular object by filling in the blanks. If the area through which heat is transferred is increased by a factor of 2, then the rate of heat transfer is increased by a factor of 2.

If the thickness of the material through which heat is transferred is increased by a factor of 2, then the rate of heat transfer is decreased by a factor of 2.