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Counting Calories
Counting Calories
A calorie is the amount of energy needed to raise the temperature of 1g of water 1°C. Using our heat equation and the specific heat of water:
Q = mCpΔT
1 calorie = (1g)(4.184 J/g°C)(1°C)
1 calorie = 4.184 J
You may be familiar with the Calorie count on food labels or packaging. This number is actually a kilocalorie, or 1000 calories.
We can measure the amount of energy released from a sample of food by burning it in a insolated device called a bomb calorimeter.
If we assume that all of the heat from burning the sample goes into the water then we can find the energy released by using our heat equation: Q = mCpΔT
However, most Calories from food are not measured directly, but instead estimated using the Atwater system of 4 Calories per gram of protein, 9 Calories per gram of fat, and 4 Calories per gram of carbohydrates.
calorimetry calculations
calorimetry calculations
In calorimetry, we assume that all of the energy released by our sample is absorbed by the water in the calorimeter. This isn't strictly true. Some energy is absorbed by the calorimeter itself and by the atmosphere. But with a well insulated calorimeter the amount of energy lost by the system can be negligible. Therefore, we assume:
- Qsample = Qwater
Notice the negative sign. The reason for the opposite sign is one half of the equation is exothermic (releases energy) and one half is endothermic (absorbs energy)
Example 1
Example 1
You drop a hot piece of copper into a calorimeter. The copper weighs 10.0g and began at 100°C. There is 100.g of water in the calorimeter and it began at 24.0°C and ends at 24.7°C. The specific heat of water = 4.184 J/g°C. What is the specific heat of copper?
We begin by assuming:
- Qsample = Qwater
Therefore:
- mlcopper Cpcopper ΔTcopper = mwater Cpwater ΔTwater
Using the information from the problem:
- (10.0g)(Cpcopper)(24.7-100°C) = (100.g)(4.184J/g°C)(24.7-24.0°C)
- (10.0g)(Cpcopper)(-75.3°C) = (100.g)(4.184J/g°C)(0.7°C)
Rearrange using algebra:
Cpcopper = (100.g)(4.184J/g°C)(0.7°C)/[- (10.0g)(-75.3°C)]
Cpcopper= 0.389 J/g°C
Example 2
Example 2
You place a hot piece of lead into a calorimeter. The lead weighs 50.0g and began at 300.°C. The calorimeter is filled with 200.g of water, which began at 24.0°C. The specific heat of water = 4.184 J/g°C. the Specific heat of lead = 0.129 J/g°C. What is the final temp in the calorimeter?
First we assume that the heat released by the lead all transfers into the water.
- Qlead = Qwater
Then using the heat equation fill in information from the problem:
- mlead Cplead ΔTlead = mwater Cpwater ΔTwater
- (50.0g)(0.129 J/g°C)(Tf - 300°C) = (200.g)(4.184 J/g°C)(Tf - 24.0°C)
Heat will continue to transfer until the water and lead reach the same temperature, so we know that the final temperature has to be the same. Let simplify the equation using some algebra and solve for the final temp.
Multiply the parenthesis and cancel units:
(-6.45J/°C)Tf + 1935J = (836.8J/°C)Tf - 20,083.2J
Move final temp to one side and combine like terms:
22,018.2J = (843.25/°C)Tf
Solve by division: 22,018.2J/(843.25J/°C) = Tf
26.1°C = Tf
Practice
Practice
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