Course Icon

Physics

Chapter 14: 5-8

SO Icon

Web Lecture

Conduction, Convection, and Radiation

Introduction

I let freeze water in drinking glasses in which I had put thermometers whose bulbs thereby were enclosed by the ice. Thereafter I brought this little apparatus to a fire. The thermometers rose until the very moment as the ice started to melt. All fire that intruded the ice thereafter ceased to have an effect on the thermometers, because it was used to make water.

— Jean André Deluc Recherches sur les modifications de l'atmosphère1787

Outline

Heat Transfer Methods

In order to determine exactly how heat moves from one object to another, we need accurate methods of measuring heat content before, during, and after a transfer activity.

Calorimetry

Obviously, a simple straightforward method of measuring heat flow would be of vast importance in studying heat phenomena. This is generally accomplished using a calorimeter, a device which isolates a system from the outside and measures changes (there is a pretty good description of a simple calorimeter and its use in this week's lab).

Example: You heat 0.150 kg of copper to 100 °C and then pour it into your calorimeter, which contains 0.200kg of water at 20°C. The final temperature of the mixture in the cup is 25°C. If the aluminum cup has a mass of 0.037 kg, what is the specific heat of copper?

What information are you given?
  • the mass of the metal mc 0.150 kg
  • the mass of the water mw 0.200 kg
  • the mass of the aluminum calorimeter cup ma 0.037
  • the specific heat of water cw (check tables) 4186 J/kgC° = 1.0 kcal/kgC°
  • the specific heat of aluminum ca (check tables ) 920 J/kgC° = 0.22 kcal/kgC°
  • the temperature of the copper T c 100 °C
  • the initial temperature of the calorimeter T i 20°C
  • the final temperature of the system after T f 20°C
What do you need to find?
  • the specific heat of copper cc
What relationships do we know?
  • heat = Q = m * c * δT
  • Q (lost by copper) = Q (gained by water) + Q (gained by cup)
  • Qc (lost by copper) = mc * cc * (T c - T f)
  • Qw (gained by water) = mw * cw * (T i - T f)
  • Qa (gained by aluminum) = ma * ca * (T i - T f)
  • heat lost = heat gained, so Qc = Qw + Qa =
  • mc * cc * (T c - T f) = mw * cw * (T f - T i) + ma * ca * (T f - T i)
Check UNITS! Our information on temperature is in Centigrade; the formula requires Kelvin, but all the temperatures are differences and the degrees are the same size, so don't bother to convert!
Now we are ready to put it all together Solve for cc = (mw * cw * (T f - T i) + ma * ca * (T f - T i)) / mc * (T c - T f)

= [( 0.200 kg * 1.00 kg/JC° * (25 -20 °C) + 0.037 * 0.22 kcal/JC° * (25 - 20 °C) / 0.150 kg (100 - 25°C)]

= 0.093 kcal/kg C° = cc

Latent Heat

When matter changes phases, all the heat input goes into converting the matter from one phase to another. During the phase change, the temperature of the material does not rise or fall (except locally). This heat flow is called latent heat, and is represented by the formula Q = m * l, where l is like specific heat, dependent on the type of material and the type of phase change (solid to liquid, liquid to gas).

Suppose we have a situation where we are converting a 0.30 kg piece of ice at -10°C to water at room temperature (20 °C). How do we represent the heat flow?

From conservation of energy, heat input to system = heat used to raise ice to 0°C + heat used to melt ice + heat used to raise water from 0°C to 20°C, or

Q(in) = m * cw * (0 - (-10)) + m * lw + m * cw * (20 - 0).

Always take into account any phase changes in the same way. Once you have the relationship between heat input and heat used to change temperatures or support phase changes, you can use the method in the calorimetry section to relate known quantities and solve for unknowns.

Heat Transfer

heat types

Three methods of heat transfer are common: conduction, convection, and radiation.

Conduction involves actually transferring kinetic energy from one system to another, and can be measured as a rate of heat flow δQ/δt = k * A * δT /l, where Q is the heat, t is elapsed time, k is a constant called the thermal conductivity of the matter involved, A is the area through which the heat is flowing, δT is the temperature difference on either side of the material transmitting the heat, and l is the length through which the heat must flow.

Obviously (from looking at the equation, and thinking about the situation), heat flow will increase if we increase the area through which it can flow or the difference in temperature. For example, (assuming that it is colder outside than in) more heat will flow out of a larger window than a smaller one, and more heat will flow if the temperature inside is 20°C and the temperature outside is 0°C than if the outside temperature is a nice 18°C. Also, if we increase the amount of material through which the heat must flow (by making the glass thicker, which is equivalent to increasing l), we can cut down on heat loss.

Convection involves moving heated matter from one physical location to another--like a hot air furnace or an air conditioner.

Radiation is the transmission of energy by electromagnetic means (i.e., light). The amount of heat a body can radiate or absorb is a function of its emissivity (close to 1 for good radiators, close to 0 for bad radiators, 0 < e < 1), the surface area through which the energy radiates, and the fourth power of the temperature of the body: P (power = energy/time) = sigma * A * e * T4. The sigma is the Stefan-Boltzmann constant, 5.67 * 10-8 watts/meters 2* K4.

Practice with the Concepts

When stuck in the wilderness during a raging storm, mycroft covered himself with snow. Why?

Discussion Points