ch18-p037 - Q = L F m = (109 kJ/kg) (0.510 kg) = 55.59 kJ...

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37. To accomplish the phase change at 78°C, Q = L V m = (879 kJ/kg) (0.510 kg) = 448.29 kJ must be removed. To cool the liquid to –114°C, Q = cm | Δ T | = (2.43 kJ/ kg K ) (0.510 kg) (192 K) = 237.95 kJ, must be removed. Finally, to accomplish the phase change at –114°C,
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Unformatted text preview: Q = L F m = (109 kJ/kg) (0.510 kg) = 55.59 kJ must be removed. The grand total of heat removed is therefore (448.29 + 237.95 + 55.59) kJ = 742 kJ....
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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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