ch18-p032 - Now following the same procedure as shown in...

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32. We note that the heat capacity of sample B is given by the reciprocal of the slope of the line in Figure 18-32(b) (compare with Eq. 18-14). Since the reciprocal of that slope is 16/4 = 4 kJ/kg·C°, then c B = 4000 J/kg·C° = 4000 J/kg·K (since a change in Celsius is equivalent to a change in Kelvins).
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Unformatted text preview: Now, following the same procedure as shown in Sample Problem 18-4, we find c A m A ( T f − T A ) + c B m B ( T f − T B ) = 0 c A (5.0 kg)(40°C – 100°C) + (4000 J/kg·C°)(1.5 kg)(40°C – 20°C) = 0 which leads to c A = 4.0×10 2 J/kg·K....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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