Question 4.18: Specific Heat and Temperature Change If we add 450. cal of h...
Specific Heat and Temperature Change
If we add 450. cal of heat to 37 g of ethanol at 20.°C, what is the final temperature?
Strategy
The equation we have has a term for temperature change. We use the information we are given to calculate that change. We then use the value we are given for the initial temperature and the change to find the final temperature.
The specific heat of ethanol is 0.59 cal/g · °C (see Table 4.2).
| Table 4.2 Specific Heats for Some Common Substances | |||
| Specific Heat (cal/g · °C) | Substance | Specific Heat (cal/g · °C) | Substance |
| 0.42 | Wood (typical) | 1.00 | Water |
| 0.22 | Glass (typical) | 0.48 | Ice |
| 0.20 | Rock (typical) | 0.48 | Steam |
| 0.59 | Ethanol | 0.11 | Iron |
| 0.61 | Methanol | 0.22 | Aluminum |
| 0.56 | Ether | 0.092 | Copper |
| 0.21 | Carbon tetrachloride | 0.031 | Lead |
Amount of heat = \text{SH} \times m \times \Delta T
Amount of heat = \text{SH} \times m \times (T_{2}-T_{1})\\ 450.\ \text{cal} = 0.59\ \text{cal/g}·°\text{C} \times 37\ \text{g} \times (T_{2}-T_{1})
We can show the units in fraction form by rewriting this equation.
450.\ \text{cal} = 0.59 \frac{\text{cal}}{\text{g}·°\text{C}} \times 37\ \text{g} \times (T_{2}-T_{1})\\ (T_{2}-T_{1}) = \frac{\text{amount of heat}}{\text{SH} \times m}\\ (T_{2}-T_{1}) = \frac{450.\ \cancel{\text{cal}}}{\left[\frac{0.59\ \cancel{\text{cal}}\times 37\ \cancel{\text{g}}}{\cancel{\text{g}}·°\text{C}} \right]} = \frac{21}{1/°\text{C}} = 21°\text{C}(Note that we have the reciprocal of temperature in the denominator, which gives us temperature in the numerator. The answer has units of degrees Celsius). Because the starting temperature is 20°C, the final temperature is 41°C.
Is this answer reasonable? The specific heat of ethanol is 0.59 cal/g·°C. This value is close to 0.5, meaning that about half a calorie will raise the temperature of 1 g by 1°C. However, 37 g of ethanol need approximately 40 times as many calories for a rise, and 40 \times \frac{1}{2} = 20 calories. We are adding 450. calories, which is about 20 times as much. Thus, we expect the temperature to rise by about 20°C, from 20°C to 40°C. The actual answer, 41°C, is quite reasonable.
■ Quick Check 4.18
A 100 g piece of iron at 25°C is heated by adding 230. cal. What will be the final temperature? Check your answer to see whether it is reasonable.