Specific Heat

# Figure 2 shows two systems thermally interacting with

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Unformatted text preview: bly put an ice cube into a hot drink to cool it quickly. You were engages, in a somewhat trial-and-error way, in a practical aspect of heat transfer known as calorimetry. Figure 2 shows two systems thermally interacting with each other but isolated from System 1 everything else. Suppose they start at different temperatures T1 and T2. As you know from Qnet = Q1 + Q2 = 0 T1 experience, eat energy will be transferred from the hotter to the colder system until they reach a common final temperature Tf. The systems will Q1 then be in thermal equilibrium and the Q2 System 2 temperature will not change further. The insulation prevents any heat energy T2 from being transferred to or from the environment, so energy conservation tells us that any energy leaving the hotter system must enter the colder system. That is, the systems exchange energy with no net less or gain. The concept is straightforward, but to state the idea Figure 2 Two systems interacting thermally. mathematically we need to be careful with signs. Let Q1 be the energy transferred to system 1 as heat. Q1 is positive if energy enters system 1, negative if energy leaves system 1. Similarly, Q2 is the energy transferred to system 2. The fact that the systems are merely exchanging energy can be written as | | | | (11) That is, the energy lost by the hotter system is the energy gained by the colder system. But Q1 and Q2 have opposite signs, so (12) Because no energy is exchanged with the environment, it makes more sense to write this relationship as 0 6 (13) This idea is not limited to the interaction of only two systems. If three or more systems are combined in isolation from the rest of their environment, each at a different initial temperature, they will all come to a common final temperature that can be found from the relationship 0 (14) NOTE: The signs are very important in calorimetry problems. ΔT is always Tf -Ti, so ΔT and Q are negative for any system whose temperature decreases. The proper sign of Q for any phase change must be supplied by you, depending on the direction of the phase change. 2.8 Problem Solving Strategy Model Visualize Solve • • • Assess Identify the interacting systems. Assume that they are isolated from the larger environment. List known information and identify what you need to find. Convert all quantities to SI units. The mathematical representation, which is a statement of energy conservation, is Qnet = Q1 + Q2 + Q3 + Q4 + L = 0 For systems that undergo a temperature change, Q = Mc(Tf-Ti). Be sure to have the temperatures in the correct order. For systems that undergo a phase change, Q = ±ML. Supply the correct sign by observing whether energy enters or leaves the system during the transition. Some systems may undergo a temperature change and a phase change. Treat the changes separately. The heat energy is Q = QΔT+Qphase. Is the final temperature in the middle? Tf that is higher or lower than all temperatures is an indication that something is wrong, usually a sign error. 3.0 References • • • An Introduction to Thermal Physics by Daniel V. Schroeder. Addison Wesley Publishing Fundamentals of Physics extended 7th edition by Halliday, Resnick, and Walker. Wiley Publishing Physics for Scientists and Engineers: A Strategic Approach with Modern Physics by Randall D. Knight. Addison Wesley Publishing. 4.0 The Experiment The experiment that you will perform today is measuring the specific heat of a metal using a calorimeter. We start with the fact the specific heat of a metal is given by (15) ∆ where m is the mass of the sample, Q is the heat of the sample, and ΔT is the change in the temperature 7 measured in Celsius or Kelvin. In the experiment that you will perform today, there are three different systems: the water, shot, and cup. The shot will be heated up by placing it in boiling water. Thus the shot will be gaining heat. Using equation (4), we can write the equation for the heat loss as (16) Once the shot is heated, you will place it inside a cup and cool water. Thus the heat will be transferred to the cup and water and the shot will be losing heat. We know from equation (4) that the energy gained is (17) Using the conservation of energy, equation (12), (18) (19) (20) 8...
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## This document was uploaded on 03/20/2014 for the course PHYS 215 at Lafayette.

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