2. An iron ball of mass 500 g at a temperature of500 ° C is cooled by placing it

Question

2. An iron ball of mass 500 g at a temperature of500 °C is cooled by placing it ona block of ice taken

from the freezer. If the mass of the ice is800 g and its initial temperatureis 10°C, then what isthe

final configuration? That is, is the final state of theH2O all ice, ice-water mixture, water only,watersteam

mixture, or steam only? If it is a mixture, determine theproportions. Find the final temperature.

The specific heat of iron is 0.107cal/g•°C.

Hint:determine the energy required for the H2O to bring itto each end of a phase transition, and

calculate the energy available from the iron foreach of these points.

Explanation / Answer

If the iron is cooled all the way to zero, itreleases... .    mcT = 500*0.107 * 500 = 26750 calories . To warm up the ice requires... .       mcT = 800 * 0.50 *10 = 4000 calories . So all of the ice will be warmed up to zero. . Then to melt the ice requres... .         m L = 800 * 79.7 =    63760 calories...   there is not enough. . Instead we use the remaining   26750 - 4000 = 22750 calories to melt some of the ice .    mass = heat / L = 22750 /79.7 =    285.4 grams of ice are melted. . So the final stateis     500 g of iron,   285.4g of water,   514.6 g of ice,   all atzero degrees Celsius

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