Homework Helpers: Physics

9 Heat and Thermodynamics


Lesson 9–2: Heat Transfer

As with other forms of energy, thermal energy is neither created nor destroyed; it only changes form. When an object is heated, it must get that energy from somewhere, and when an object cools, the heat must go somewhere else. Another important aspect of heat transfer, based on the second law of thermodynamics, is that heat will be transferred from a hotter object to a colder object, not the other way around.

Heat can be transferred by one or by a combination of more than one of the following methods.

Conduction involves the transfer of heat between two objects that are in contact with each other. Molecules with relatively high kinetic energy crash into molecules with relatively low kinetic energy and some of their kinetic energy in the collision. When you immerse your body in hot bathwater, the water molecules strike your body and transfer energy through conduction.

Convection involves the transfer of energy along with a transfer of molecules with relatively high kinetic energy. If you have ever been in a bathtub in which the water has become uncomfortably cold, you may have turned on the faucet to release more hot water into the tub. The heat reaches you faster than could be expected if the kinetic energy had to transfer from molecule to molecule all the way from the source to your body. Currents of hot water will travel through the tub, and hot water will make its way to you. Convection also takes place with gas molecules, as when hot air is circulated through a room.

Radiation involves the transfer of heat in the form of electromagnetic waves. Much of the heat that reaches us from the sun is in this form. Microwave ovens are good examples of devices that heat objects via radiation, as are the “hot lamps” that are used to keep food warm in many cafeterias.

So, now you know the methods of heat transfer, and you know that heat is transferred from the hotter object to the colder object, but how much heat does a hot object transfer when it comes in contact with a colder object? Given time, two objects in contact with each other will eventually reach thermal equilibrium, when they both have the same temperature. The temperature at which that will occur depends on a few factors, including the masses (m) and the specific heats (C) of the objects in question. You probably already have a good understanding of mass and temperature, but let’s talk more about the concept of specific heat, or, as it is often called, specific heat capacity.

Imagine placing two spoons of equal mass, one made of wood and one made of metal, in a pot of boiling water for one minute, which is long enough to transfer some heat, but not enough to bring the temperature of the spoons up to the temperature of the water. When you take the spoons out, would you expect them to be at the same temperature, or would the metal spoon be hotter than the wooden one? If you guessed the latter scenario, you are right, and you probably have some understanding of the concept of specific heat.

The specific heat of a substance is defined as the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. Some objects will experience a greater change in temperature than others, even when the amount of heat transferred and the mass of the individual objects are the same. If you have ever gone to the beach, you probably noticed a large difference between the temperature of the sand and the water. Often, the sand is very hot, yet the water can be cool or even cold, despite the fact that they are both under the same sun. One reason for this has to do with the specific heat of each of the different substances. The specific heat of water, 4,190 J/kg × °C, or 1.00 cal/g × °C, is quite high. As a result, the temperature of water changes more slowly than the temperature of many other substances.

Heat Transfer

Q = mCΔT

heat transfer = mass × specific heat × change in temperature

Example 1

How many joules of energy are required to raise the temperature of 2.50 kg of water from 11.0°C to 33.0°C?

Given: m = 2.50 kg       ΔT = Tf – Ti = 33.0°C – 11.0°C = 2.20°C C = 4,190 J/kg · °C

Find: Q


Notice how the units cancel out in Example 1. Make sure that if you are given a problem that deals with calories, you make the appropriate conversions or select the specific heat in terms of calories.

Example 2

A 3.50 kg sample of water with an initial temperature of 17.0°C absorbs 8,840 calories of heat from its surroundings. What is the final temperature of the water?

Note: The heat transferred (Q) is given in calories, so I will select the specific heat of water in calories as one of my givens. Alternatively, I could use the conversion factor, 1 cal = 4.19 J, to convert calories to joules.

Given: m = 3.50 kg     Ti = 17.0°     Q = 8,840 cal     C = 1.00 × 103 cal/kg × °C

Find: Tf

Lesson 9–2 Review

1. _______________ is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.

2. Calculate the specific heat capacity of an unknown material if it takes 33,400 J of energy to raise the temperature of a 1.3 kg sample of the substance by 14°C.

3. How much heat, in joules, would be required to raise the temperature of 45.5 kg of water by 25.0°C?