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Unformatted text preview: =600 g to bring it from T i = − 10 ◦ C to T f =+25 ◦ C? • L V =2256 kJ/kg • L F =333 kJ/kg • c ice =2 . 05 kJ/kg/K • c liquid =4 . 18 kJ/kg/K • c gas =2 . 08 kJ/kg/K There are 3 successive stages of heating in this problem: (a) heating ice: Q i (b) changing phase: Q il (c) heating liquid: Q l So, the total heat required is: Q tot = Q i + Q il + Q l = mc i Δ T i + mL F + mc l Δ T l = m ( c i Δ T i + L F + c l Δ T l ) where I’ve redeFned c i = c ice and c l = c liquid . The values are predeFned. Additionally: • Δ T i =0 ◦ C − ( − 10 ◦ C )=10 ◦ C =10 K • Δ T l =25 ◦ C − ◦ C =25 ◦ C =25 K Substituting in values: Q tot = 0 . 6 kg ((2 . 05 kJ/kg/ a a K )(10 a a K ) + 333 . kJ/kg + (4 . 18 kJ/kg/ a a K )(25 a a K )) = 0 . 6 A A kg ( 20 . 5 kJ/ A A kg + 333 . kJ/ A A kg + 104 . 5 kJ/ A A kg ) = (0 . 6)(458 kJ ) ±inally: Q tot = 274 . 8 kJ 2...
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This note was uploaded on 05/01/2010 for the course PHYS 224 taught by Professor Hanlet during the Spring '10 term at Illinois Tech.
 Spring '10
 Hanlet

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