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report for T2

# report for T2 - 1 2 f i c T T M C gm cal Q-⋅ ° ⋅ =...

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LAB REPORT FOR T2 EXPERIMENT: MELTING OF ICE VI-1: For this part of the report we were required to find the initial and final temperatures of the water based on the interval we determined in our plot as well as the resistance readings for those. From my plot on the following page you can see that my initial and final resistances are as follows: R i = 65000 ohms R f = 290000 ohms The formula to find temperature from resistance is as follows for both temperatures: where R 1 and R 2 are temperatures corresponding to the table in the manual. 2 1 1 1 ) ( R R R R T R T i i - - + = - - + ° = 63480 66356 65000 66356 34 ) 65000 ( C T C C T ° + ° = 47 . 0 34 ) 65000 ( C T i ° = 47 . 34 2 1 1 2 ) ( R R R R T R T f f - - + = - - + ° = 283600 298990 290000 298990 3 ) 290000 ( C T C C T ° + ° = 58 . 0 3 ) 290000 ( C T f ° = 58 . 3 VI-2: For this part of the report I will calculate the latent heat using equation 2 from the manual: M c +M w +M I 184.2gm M c 51.3gm M w 77.3gm M I 55.6gm Heat lost by water: ) ( 1 1 f i w T T M C gm cal Q - ° = = (1cal/gm °C)(77.3gm)(34.47°C -3.58°C) = 2387.797cal Heat lost by cup:

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Unformatted text preview: ) ( 1 2 f i c T T M C gm cal Q-⋅ ° ⋅ = = (1cal/gm °C)(51.3gm)( 34.47°C -3.58°C) = 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)( 3.58°C-0) = 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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