42 Re measure diameter of ice block as d2 43 Repeat test with second material

42 re measure diameter of ice block as d2 43 repeat

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4.2 Re-measure diameter of ice block as d2 4.3 Repeat test with second material plate sample. Assume ambient melting rate is consistent. 4.4 Clean up Results Table #2 results Sample h (mm) t (min) mcw (g) mw (g) Davg(mm) A( mm 2 ) k ( W /( m×K ) ¿ Ambien t xxx 5 225.73 7.73 82.1 5294 xxx Sample 1 6.7 5 231.95 13.95 79.65 4983 12.6 × 10 4 Sample 2 6.4 5 282.48 64.48 77.3 4693 9.79 × 10 4 Discussion
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We found that the thermal conductivity of sample1 was 12.6 × 10 4 W m×K , and was 9.79 × 10 4 W m× K for sample2. Errors in this lab are human error, such as mistakes and inconsistencies measuring masses or diameters, or the precision of the steam apparatus in consistancy. The accuracy of the instruments used for measurements could be slightly off. The environment could also have an effect on final results if the stock pot got hotter/colder or there was a change in the ambient temperature of the room. Conclusion In conclusion, we found that when a block of ice is placed on a sample, it melts fastest with high temperatures, therefore with materials of a higher thermal conductivity. From our collected data, we were able to calculate the thermal conductivity of each material. The data found was consistent with theories learnt in class. Example of calculations ∆Q = kA ( T 2 T 1 ) ∆T h k = thermal conductivity ( W m×K ) A = area ( mm 2 ) ∆T = temperature ( ) h = thicknessof samplematerial ( mm ) Lw = latent heat of water ( J g )
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( mw ) ( Lw ) = ( k ) ( A ) ( Tf Ti ) ∆T h k = ( mw ) ( Lw ) ( h ) A ( Tf Ti ) ∆T k sample 1 = ( 13.97 g ) ( 334 J g ) ( 6.7 mm ) ( 4983 mm 2 ) ( 100 ° 0 ° ) ( 300 s 0 s ) ¿ 0.0012590295 W mK ¿ 12.6 × 10 4 W mK Bibliography Georgia State University. (n.d.). Specific Heat. Retrieved October 26, 2016, from - astr.gsu.edu/hbase/thermo/spht.html Introducing Static and Dynamic Mechanisms. (n.d.). Thermal Conductivity. Retrieved October 26, 2016, from
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