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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 T i + mL F + mc l T l = m ( c i T i + L F + c l T l ) where Ive redeFned c i = c ice and c l = c liquid . The values are predeFned. Additionally: T i =0 C ( 10 C )=10 C =10 K 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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 Spring '10
 Hanlet

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