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Unformatted text preview: ) ( 1 2 f i c T T M C gm cal Q = = (1cal/gm C)(51.3gm)( 34.47C 3.58C) = 348.625cal Heat gained by melting ice: I LM Q = 3 Heat gained by melting ice: ) ( 1 4 C T M C gm cal Q f I  = = (1cal/gm C)(55.6gm)( 3.58C0) = 199.048cal To solve for L we use equation 2 which is: ) ( 1 ) ( 1 ) ( 1 f i c f i w f I I T T M C gm cal T T M C gm cal T M C gm cal LM + = + Q 3 + Q 4 = Q 1 + Q 2 L(55.6gm) + 199.048cal = 2387.797cal + 348.625cal L = 45.64 cal/gm From the site http://en.wikipedia.org/wiki/Enthalpy_of_fusion I found that the accepted value of L should have been 79.72 cal/gm therefore the % difference is as follows: %difference = [(measured accepted)/accepted] x 100 %difference = [(45.64cal/gm 79.72cal/gm)/79.72cal/gm] x 100 %difference = 42.75%...
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This note was uploaded on 02/17/2009 for the course PHY 151 taught by Professor Shi during the Fall '08 term at SUNY Buffalo.
 Fall '08
 SHI
 Physics, Resistance

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