*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **kg .20 kg .069 kg 150 . = + + c) Due to the much larger quantity of ice, a reasonable initial guess is an ice-water mix at C. . ° The energy required to melt all of the ice would be = × kg) J 10 (334 kg) 350 . ( 3 J 10 17 . 1 5 × . The maximum energy that could be transferred to the ice would be if all of the steam would condense and cool to C . ° and if all of the water would cool to C . , )) C K)(100.0 kg J 4190 ( kg J 10 (2256 kg ) 0120 . ( 3 ° ⋅ + × + J. 10 6.56 ) C K)(40.0 kg J kg)(4190 200 . ( 4 × = ° ⋅ This is insufficient to melt all of the ice, so the final state of the system is an ice-water mixture at J 10 56 . 6 . C . 4 × ° of energy goes into melting the ice. So, kg J 10 334 J 10 56 . 6 3 4 × × = m kg. 196 . = So there is kg 154 . of ice, and kg 0.20 kg 0.196 kg 012 . + + kg 0.408 = of water....

View
Full
Document