MT1-cheatsheet

as usual before you start to manipulate equations

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: volume, the heat capacity is cv = 3/2 N kB Hint 2: dS = dQ/T Problem 4 (25 points). As you know, the Carnot engine is the most efficient engine running between a hot reservoir at temperature T2 and a cold reservoir at temperature T1. Unfortunately, there are no true Carnot engines, because in order to run an engine reversibly, it must also run infinitely slowly. In that case, you get no power from the engine. Let's construct an engine from which we can get some power. We will call this a "Carnot-like" engine. To be specific, assume that during the hot (or cold) part of the cycle, the engine is in contact with the hot (or cold) reservoir through a metal barrier with conductivity , area A, and width w. Assume whatever else you need. The broad questions for this problem are these: What is the maximum power that can be obtained from a Carnot-like engine running between temperatures T2 and T1, and what is the efficiency when it is running at maximum power? a) Derive an expression for the maximum power that can be delivered by this engine. Your answer should be expressed in terms of T2, T1, and = A/w. Hint: derive an expression for the power as a function of the temperature of the engine during the hot part of its cycle. b) What is the efficiency of this engine when it is delivering the maximum power? Physics 7B Fall, 2006 Homework 2 Solutions 19.13 We find the temperature from heat lost = heat gained; mshoecshoe Tshoe = (mpotcpot + mwatercwater) Tpot ; (0.40 kg)(450 J/kg C)(T 25C) = [(0.30 kg)(450 J/kg C) + (1.35 L)(1.00 kg/L)(4186 J/kg C)](25C 20C), which gives T = 186C. 19.23 The temperature of the ice will rise to 0C, at which point melting will occur, and then the resulting water will rise to the final temperature. We find the mass of the ice cube from heat lost = heat gained; (mAlcAl + mwatercwater) TAl = mice(cice Tice + Lice + cwater Twater); [(0.075 kg)(900 J/kg C) + (0.300 kg)(4186 J/kg C)](20C 17C) = mice{(2100 J/kg C)[0C ( 8.5C)] + (3.33 105 J/kg) + (4186 J/kg C)(17C 0C)}, which gives mice = 9.4 103 kg = 9.4 g. 19.26 The s...
View Full Document

Ask a homework question - tutors are online