13.3: Heat
Objectives
The student will:
- Explain the relationship between heat and energy transfer.
- Describe how the calorie is a measure of energy.
Vocabulary
- calorie: The amount of energy needed to increase the temperature of 1 gram of water by 1 °C at standard atmospheric pressure (101.325 kPa). This is approximately 4.2 joules.
- heat: The sum of all the kinetic energies of all the molecules in an object or substance.
Introduction
Heat is yet another common word that has a very specific meaning in physics. We place our hands in front of a fire and speak of heat. We step with bare feet onto a tiled floor in an unheated room during winter and our feet feel cold. We speak of “losing heat.” Human beings have expressed comfort and discomfort concerning the pleasant and unpleasant effects of heat since the advent of language. But an understanding of the nature of heat was not a simple affair. Even as late as the 18th century it was still not properly understood. For example, it was still commonly accepted that if two objects had the same temperature then they “contained” the same amount of heat. We know today that this is not true. During the latter half of the 18th century, the idea that heat was a form of matter called caloric became generally accepted. This, too, was proven incorrect. It was not until the 19th century that physicists were finally able to explain the nature of heat.
Understanding Heat
By the mid-19th century, R.J. Mayer constructed the principle of the conservation of energy. He based his premise on an experiment comparing mechanical energy and heat. Mayer’s ideas were taken up by James Prescott Joule, whose work was discussed earlier. One of Joule’s famous experiments involved heating a fluid mechanically, using friction. In Joule’s experiment, a paddle wheel inside a container of fluid was rotated, resulting in an increase of the fluid’s temperature. The rise in the temperature of the fluid indicated that the work done in rotating the paddle wheel was equivalent to heating up the fluid (with a flame, for example). Heat could therefore be viewed as a transfer of energy, involving a temperature change. Thus the definition of heat is given as:
Energy that is transferred from one object to another object due to a temperature difference between the objects.
To restate: When we speak of heat, we do not mean a material substance, nor do we mean a form of energy. Again, by heat we mean a transfer of energy.
The Calorie
The amount of energy required to raise the temperature of one gram of water by one degree Celsius is defined as one calorie, or, as it is sometimes referred to, one “small” calorie. The calorie is therefore a measure of energy. It is equal to about 4.186 J. Using the definition above, the caloric content of the average banana is 80,000 calories. A more useful unit to measure food calories is called the “big” calorie, which equals 1000 “small” calories. Thus, the average banana contains 80 “big” calories, or 80 food calories.
\begin{align*}1000 \ small \ calories =1 \ food \ calorie=4186J \rightarrow 1.00kcal=1.00kC=4186J\end{align*}.
Check Your Understanding
Your doctor recommends an average caloric intake of 2,000 calories per day.
How many joules is this?
Answer: Since 1 food calorie equals 4186 J: \begin{align*}(2000 \ C)(4186 \frac{J}{C})=8,372,000 \rightarrow 8.372 \times 10^6 \ J\end{align*}
Illustrative Example 13.1.1
a. How long can a 60 W light bulb burn if supplied with \begin{align*}8.372 \times 10^6 \ J\end{align*}?
Answer: 60 W is the same as \begin{align*}60 \frac{J}{s}\end{align*}. Thus ,a 60 W lightbulb uses 60 J every second.
If \begin{align*}8.372 \times 10^6 J\end{align*} are available, then \begin{align*}\frac{8.372 \times 10^6 J}{60 \frac{J}{s}}=139,533 \rightarrow 140,000s\end{align*}. Since there are 86,400 seconds in one day, \begin{align*}\frac{139,533}{86,400}=1.6149 \rightarrow 1.61 \ days\end{align*}, or close to 39 hours.
b. To what height could a 70.0 kg person climb using this much energy? See Figure below
A vertical climb
Answer: We know that 2,000 kcal equals \begin{align*}8.372 \times 10^6 J\end{align*}. So, 20% of this is \begin{align*}0.20 \times 8.372 \times 10^6=1.674 \times 10^6 \ J\end{align*}
Recall that the net work done by a climber of weight mg over a vertical displacement of \begin{align*}\Delta h\end{align*} is equal to the change in his potential energy:
\begin{align*}\Delta PE=mg \Delta h\end{align*} (Assuming the change in his kinetic energy is zero.)
Therefore:
\begin{align*}mg \Delta h=1.674 \times 10^6 \rightarrow (70.0kg)\left(9.81 \frac{m}{s^2}\right) \Delta h=1.674 \times 10^6 \rightarrow\end{align*}
\begin{align*}\Delta h=\frac{1.674 \times 10^6}{(70.0kg)\left(9.81 \frac{m}{s^2}\right)}=2437.75 \rightarrow 2,440m!\end{align*}
Most of the calories (energy) that we consume are used by our bodies even if we do not exercise or otherwise exert ourselves physically. For example, they are used to maintain a constant body temperature and keep our heart pumping. It turns out that about 80% of our caloric intake is used just to keep us alive. The remainder of the energy is used to perform work, such as climbing a flight of stairs, taking a walk, eating, talking, or anything else that requires you to use energy in your daily routine. If you use more than the remaining 20% of the energy you’ll lose weight, since your body uses your stored fat to make up the difference. On the other hand, if you don’t use up the remaining 20%, the excess is converted to fat and you’ll gain weight.
Illustrative Example 13.2.2
a. A dieting person decides to eliminate the cream from his coffee. If he typically drinks two cups of coffee per day with one serving of cream added to each cup, how many calories would be saved over the course of a year? One serving of cream (about 15 grams, or one tablespoon) contains about 20 calories.
Answer:
\begin{align*}40 \frac{Cal}{day} \times 365 \ days=14,600 \ Cal\end{align*}
b. Assuming the person’s metabolic rate remains fixed while dieting, (highly unlikely, actually) what maximum weight loss can be expected? One kilogram of fat contains about 8,800 calories (one pound of fat has about 4000 calories).
From part A we know that 14,600 cal were saved, therefore
\begin{align*}\frac{14,600 \ Cal}{8,800 \frac{Cal}{kg}}=1.659 \rightarrow 1.66 \ kg \ (3.65 \ pounds)\end{align*}.
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