14_InstSolManual_PDF_Part8

# 14_InstSolManual_PDF_Part8 - 14-8 Chapter 14 For the water,...

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For the water, gives Reflect: We assumed that the water remained liquid. This is a correct assumption since the final temperature we cal- culated is in the range 14.46. Set Up: For water, For copper, Solve: For the water, For the copper can, For the sample, gives 14.47. Set Up: Since some ice remains, the ice and water from the melted ice remain at For silver, For water, Solve: For the silver, For the ice, where m is the mass that melts. gives and Reflect: The heat that comes out of the ingot when it cools goes into the ice to produce a phase change. 14.48. Set Up: For ice, For water, and Solve: For the water, For the ice, gives and 14.49. Set Up: Ice remains, so the final temperature is For water, Solve: For the sample, For the vial, For the ice that melts, gives and 14.50. Set Up: Solve: 14.51. Set Up: The thermal gradient is Solve for and then where We want Q for each square meter, so set A 5 1.0 m 2 . t 5 1 day 5 8.64 3 10 4 s. Q 5 Ht , H 5 Q / t T H 2 T C L 5 1 25 K / km 2 1 1 km 10 3 m 2 5 0.025 K / m. Q 5 kA 1 T H 2 T C 2 t L 5 1 0.075 W / m # K 21 100 3 10 2 4 m 2 21 80 C° 21 86,400 s 2 3.00 3 10 2 2 m 5 1.73 3 10 5 J 1 day 5 86,400 s. m 5 3.08 3 10 2 3 kg 5 3.08 g. mL f 2 702 J 2 328 J 5 0 g Q 5 0 Q i 5 mL f . Q v
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## This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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