T = q / C = (-5.00x103 J) / (2.50x103 J K-1) = -2.00K, or –2.00oC. QUESTION: What is the final temperature of an object, initially at 25.0oC, if 5.00x103 J of heat flows out of it? Assume that the heat capacity is 2.50x103 J K-1. A.–250.0oC, B. –2.00oC, C. 23.0oC, D. 27.0oC, Assuming no chemical or phase change, the temperature change, T, is directly proportional to the amount of heat flow: q = C T where the proportionality constant, C, is the heat capacity. Therefore: T = q / C = (-5.00x103 J) / (2.50x103 J K-1) = -2.00K, or –2.00oC. Tfinal = Tinitial + T = (25.0oC) + (-2.00oC) = 23.0oC What is the final temperature of an object, initially at 25.0oC, if 5.00x103 J of heat flows out of it? Assume that the heat capacity is 2.50x103 J K-1. A.–250.0oC, B. –2.00oC, C. 23.0oC, D. 27.0oC, PAUSE First of all, if heat flows out of the system, we expect the temperature of the system to drop, as long as no chemical or phase changes are happening. Therefore, we can immediately rule out choice D. The initial temperature is 25 degrees celsius HIGHLIGHT 25.0oC And the answer given in choice C is 27 degrees…. Which is higher than the initial temperature. But how do we figure out which of choices A, B and C is correct? CLICK For as long as no chemical or phase changes are happening, the amount of heat flow, which is often represented by the letter q, HIGHLIGHT q in q = C T is directly proportional to the temperature change. HIGHLIGHT T in q = C T Mathematically, we express this idea by saying that q is equal to a constant times delta T. The constant is shown here as C, HIGHLIGHT C and is called the proportionality constant. In this particular case, the proportionality constant is referred to as the heat capacity. Let’s examine what we’re given in this problem. We are given the heat capacity… HIGHLIGHT value of heat capacity in problem and C in equation (q = C T) --- same color We are also given the amount of heat flow, HIGHLIGHT 5.00x103 J in problem Since we are given that heat flows out of the system, our value for q is a negative number. CALLOUT “q = - 5.00x103 J” pointing to q in equation Since we know C and q, we can solve for the temperature change. CLICK CLICK delta T, is equal to q HIGHLIGHT q and (-5.00x103 J) divided by C. HIGHLIGHT C and (2.50x103 J K-1) Carrying out the indicated calculation gives us a value of negative 2.00 kelvin, or negative 2.00 degrees Celsius. A negative value of delta T means a drop in temperature, as expected. Note that a temperature change of one kelvin corresponds to a one degree change in the Celsius reading. We do not add or subtract 273 when converting temperature changes between celsius and kelvin. We express our calculated temperature change to 3 significant digits. CALLOUT “3 sig.figs.” pointing to HIGHLIGHTED 2.00 in 2.00K because both terms we used in the calculation have 3 signifciant digits CALLOUT “3 sig.figs.” pointing to “5.00x103 J” CALLOUT “3 sig.figs.” pointing to “2.50x103 J” Before you rush to say that choice B is the correct answer, let’s re-read the question. We are being asked for the final temperature, not the temperature change. The answer given in choice B, negative 2.00 degrees Celsius, is the temperature change. To get the final temperature, we add the temperature change to the initial temperature. The initial temperature is 25.0 degrees Celsius HIGHLIGHT 25.0oC and Tinitial, same color Adding to that the change in temperature, which is –2.00 degrees celsius, HIGHLIGHT (–2.00oC) Gives us a final temperature of 23.0 degrees celsius. Therefore, the correct answer is C. Note that our initial temperature, 25.0, is given to the nearest tenth of a degree. And we calculated delta T to the nearest hundredth of a degree. Therefore, our answer should only be expressed to the nearest tenth of a degree. CLICK PAUSE END RECORDING

Video ID: 8-3-3 © 2008, Project VALUE (Video Assessment Library for Undergraduate Education), Department of Physical Sciences Nicholls State University Author: Glenn V. Lo Assisted by Johuan Gasery In partial fulfillment of the requirements for CHEM 481, Fall 2008.