Specific Heat Capacity & Calorimetry
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- Garry Doyle
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1 Unit: Thermal Physics Knowledge/Understanding: specific heat capacity calorimetry Skills: solve calorimetry (specific heat) problems Notes: Different objects have different abilities to hold heat. For example, if you enjoy eggplant Parmesan, you may have noticed that the eggplant holds much more heat (and burns your mouth much more readily) than the cheese or the bread crumb coating. The amount of heat that a given mass of a substance can hold is its specific heat capacity. specific heat capacity: a measure of the amount of heat required per gram of a substance to produce a specific temperature change in the substance. (Note that specific heat capacity is usually given in kilojoules per kilogram per Kelvin (or degree elsius). Be careful with this in most other equations, energy is measured in joules.) p : specific heat capacity, measured at constant pressure. For gases, this means the measurement was taken allowing the gas to expand as it was heated.
2 v : specific heat capacity, measured at constant volume. For gases, this means the measurement was made in a sealed container, allowing the pressure to rise as the gas was heated. For solids and liquids, p v because the pressure and volume change very little as they are heated. For gases, p > v (always). For ideal gases, p v = R, where R is the gas constant. (We will see the gas constant again when we study fluid mechanics.) When there is a choice, p is more commonly used than v because it is easier to measure. When dealing with solids and liquids, most physicists just use for specific heat capacity and don t worry about the distinction. alculating Heat from a Temperature hange The amount of heat gained or lost when an object changes temperature is given by the equation: where: Q = heat (J) Q m m = mass (kg) = specific heat capacity ) ( kg K ΔT = temperature change (K or ) Again, remember that the traditional units for are rather than J. You may sometimes see the units for given in Note that 1 J 1 1 kgk kg g J cal or. g g and 1 cal kcal g kg kg K
3 Specific Heat apacities of Some Substances Substance Specific Heat apacity ) ( kg K water at ethylene glycol (anti-freeze) ice at steam at steam at vegetable oil air aluminum glass iron copper brass silver lead gold 0.129
4 alorimetry calorimetry: the measurement of heat flow In a calorimetry experiment, heat flow is calculated by measuring the mass and temperature change of an object and applying the specific heat capacity equation. calorimeter: an insulated container for performing calorimetry experiments. coffee cup calorimeter: a calorimeter that is only an insulated container it does not include a thermal mass (such as a mass of water). It is usually made of styrofoam, and is often nothing more than a styrofoam coffee cup. bomb calorimeter: a calorimeter for measuring the heat produced by a chemical reaction. A bomb calorimeter is a double-wall metal container with water between the layers of metal. The heat from the chemical reaction makes the temperature of the water increase. Because the mass and specific heat of the calorimeter (water and metal) are known, the heat produced by the reaction can be calculated from the increase in temperature of the water.
5 Solving offee up alorimetry Problems Most coffee cup calorimetry problems involve placing a object in contact with a er one. Many of them involve placing a piece of metal into water. To solve the problems, assume that both objects end up at the same temperature. The heat lost by the object equals the heat gained by the object. (However, remember that Q will be negative for the object because it is losing heat.) m Q Q m Q T m Q m Notice that there are six quantities that you need: the two masses, the two specific heat capacities (), and the two temperature changes (ΔT ). (You might be given initial and temperatures for either or both, in which case you ll need to subtract.) The problem will give you five of the six, and you will need to find the missing one. Don t fret about the negative sign. The value of ΔT will be negative for the object (because it is cooling off), and the two minus signs will cancel.
6 Steps for Solving offee up alorimetry Problems 1. Identify the variables for both the and substance. This can be tricky because you have two different masses, two different specific heat capacities, and two different temperature changes. For each quantity, you have to identify both the variable and which substance it applies to. 2. Look up the specific heat capacities of the substances involved. 3. Plug each set of numbers into the equation Q m. (I.e., you ll have two separate Q m equations.) a. Remember that for the substance that is cooling off, heat is going out of the system, which means the equation will be Q m. b. Because T will be negative for the substance that was cooling off, the two negative signs will cancel. 4. Use the fact that Q is the same for both equations to solve for the unknown quantity. This will involve doing one of the following: a. alculate the value of Q from one equation and use it in the other equation. b. If you need to find the temperature, set the two m T expressions (or m T f T ) expressions) equal to each other. ( i
7 Sample Problems: Specific Heat apacity & alorimetry Q: An kg block of aluminum is heated and placed in a calorimeter containing kg of water at 20. If the temperature of the water was 30, to what temperature was the aluminum heated? A: The heat gained by the water equals the heat lost by the aluminum. The heat gained by the water is: Q m Q (0.100 kg)(4.18 )( 10 ) kg K Q 4.18 The heat lost by the metal must therefore be (0.050 kg)(0.900 ) kg K 4.18 (0.045)( ) Q m The temperature of the aluminum was 93 (i.e., it went down by 93 ) T 123 This means the initial temperature must have been 123. i f i i
8 Q: An kg block of copper at 95 is dropped into a calorimeter containing kg of water at 25. What is the temperature? A: Once again, the heat lost by the copper equals the heat gained by the water. m (0.025 kg)(0.385 copper kgk ( T )( T T T copper ( )( T Q ΔT copper copper Q m 95) (0.3138)( T ) T T T ΔT 95 ) (0.075 kg)( T T water water water water 25) kgk )( T 25 ) Note that because the specific heat of the water is so much higher than that of copper, and because the mass of the water was larger than the mass of the copper, the temperature ended up much closer to the initial water temperature.
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