Sp05-MT1-Liphardt-So - Lecture 1 Midterm 1 Solutions Spring 2005 Physics 7B Midterm 1 Lecture 1 Problem 2 Solution NOTE Heat capacities were given

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Physics 7B Midterm 1 Lecture 1 Problem 2 Solution NOTE: Heat capacities were given in this problem not specific heats (Heat Capacity = Mass x Specific Heat) This is why we did not need masses. A simple check of the units would have revealed this. Part a We have to integrate because the heat capacity is a function of temperature. dQ = c L dT Q = c L dT = aT 3 dT T i T f = 1 4 a ( T f 4 T i 4 ) = 1 4 (0.128)(0.7 4 1 4 ) = 0.0243 J Part b Q = 0 = Q solid + Q liquid 0 = c s dT + T 0 T f c L dT T i T f = bT 2 dT + T 0 T f aT 3 dT T i T f = b ( 1 T 0 1 T f ) + a 4 ( T f 4 T i 4 ) T 0 = [ a 4 b ( T i 4 T f 4 ) + 1 T f ] 1 = [ 0.128 0.015 (1 4 0.5 4 ) + 1 0.5 ] 1 = 0.25 K Part c Δ S universe = Δ S system + Δ S environment but the system is isolated from the environment so Δ S environment = 0 ⇒Δ S universe = Δ S system Δ S system = Δ S liquid + Δ S solid = dQ liquid T + dQ solid T = c L dT T + c S dT T = aT 2 dT + bT 3 dT T 0 T f T i T f = a 3 ( T f 3 T i 3 ) + b 2 ( 1 T o 2 1 T f 2 ) = a 3 (0.5 3 1 3 ) + b 2 ( 1 0.25 2 1 0.5 2 ) = 6 b 0.29 a = 0.0529
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This note was uploaded on 02/04/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at University of California, Berkeley.

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Sp05-MT1-Liphardt-So - Lecture 1 Midterm 1 Solutions Spring 2005 Physics 7B Midterm 1 Lecture 1 Problem 2 Solution NOTE Heat capacities were given

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