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Unformatted text preview: Notes to accompany lecture on p. 70 of text. Case I. Consider a block of mass, M, falling through a height, h, ideally into a bath of water at temperature, T1. By ideally, we mean no splashing, wave generation or other energy wastes. If M = 1kg, T 1 = 298K, h = 3.0 m and C(bath) = 4.0 J/K, calculate T 2 , the final temperature of the bath. ∆ E block = Mg ∆ z = Mg ( z f z i ) = Mg (0 h ) =  Mgh ∆ E bath = q + w = q + = C bath ∆ T ∆ E tot = ∆ E block + ∆ E bath = ∆ E block = ∆ E bath C bath ∆ T =  ( Mgh ) = + Mgh ∆ T = 1.0 kg ( 29 × 9.8 m / s 2 ( 29 × 3.0 m ( 29 4 J / K = 7.4 K T 2 = 305.4 K Case II. Now block M, falling the same distance as in case I (and thereby having the same change in potential energy) now lifts a lighter block of mass, m, up a height, h (same distance through which M falls). If m = 500g = 0.50 kg, calculate T 2 . Note that we will use “sys” to denote block, M, as the “system” to distinguish it from block, m, which is part of the...
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This note was uploaded on 03/30/2008 for the course CH 201 taught by Professor Warren during the Fall '07 term at N.C. State.
 Fall '07
 Warren
 Chemistry

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