Question 2.11: If a hot-water bottle contains 1200 g of water at 65°C, how ......
If a hot-water bottle contains 1200 g of water at 65°C, how much heat, in calories, will it have supplied to a person’s “aching back” by the time it has cooled to 37°C (assuming all of the heat energy goes into the person’s back)?
We will substitute known quantities into the equation
\quad\quadHeat released = specific heat × mass × temperature change
Table 2.4 shows that the specific heat of liquid water is 1.00 cal/g · °C. The mass of the water is given as 1200 g. The temperature change in going from 65°C to 37°C is 28°C. Substituting these values into the preceding equation gives
\quad\quad\quad\quad Heat released = (\frac{1.00 cal}{\cancel{g} \cdot °\cancel{C}})\times(1200 \cancel{g})\times (28°\cancel{C})\\\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad= 33,600 cal \quad\quad (calculator answer)\\\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad= 34,000 cal \quad\quad (correct answer)
The given quantity of 1200 g and the temperature difference of 28°C, both of which have only two significant figures, limit the answer to two significant figures.
| TABLE 2.4
Specific Heats of Selected Common Substances |
|
|
Substance |
Specific Heat
(cal/g · °C)* |
| water, liquid | 1.00 |
| ethyl alcohol | 0.58 |
| olive oil | 0.47 |
| wood | 0.42 |
| aluminum | 0.21 |
| glass | 0.12 |
| silver | 0.057 |
| gold | 0.031 |
| *The unit notation cal/g · °C means calories per gram per degree Celsius. | |