problem17_p107

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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17.107: a) The possible final states are in ice-water mix at C, 0 . 0 ° a water-steam mix at C 0 . 100 ° or water at an intermediate temperature. Due to the large latent heat of vaporization, it is reasonable to make an initial guess that the final state is at C. 0 . 100 ° To check this, the energy lost by the steam if all of it were to condense would be J. 10 14 . 2 ) kg J 10 2256 )( kg 0950 . 0 ( 5 3 × = × The energy required to melt the ice and heat it to J, 10 13 . 1 )) C 100 ( K) kg J 4190 ( kg J 10 kg)(334 (0.150 is C 100 5 3 × = ° + × ° and the energy required to heat the origianl water to ) kg.K J kg)(4190 (0.200 is C 100 ° J. 10 4.19 ) C (50.0 4 × = ° Thus, some of the steam will condense, and the final state of the system wil be a water-steam mixture at C. 0 . 100 ° b) All of the ice is converted to water, so it adds 0.150 kg to the mass of water. Some of the steam condenses giving up J 10 55 . 1 3 × of energy to melt the ice and raise the temperature. Thus, kg 69 . 0 kg J 10 2256 J 10 55 . 1 3 5 = = × × m and the final mass of steam is 0.026 kg, and of the water, kg. 0.419
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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....
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