An iron ball of mass 500 g at a temperature of 500 °C is cooledby placing it on

Question

An iron ball of mass 500 g at a temperature of 500 °C is cooledby placing it on a block of ice taken

from the freezer. If the mass of the ice is 800 g and itsinitial temperature is 10 °C, then what is the

final configuration? That is, is the final state of the H2Oall ice, ice-water mixture, water only, water-

steam mixture, or steam only? If it is a mixture, determinethe proportions. Find the final temperature.

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

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

calculate the energy available from the iron for each of thesepoints.   

Explanation / Answer

mass of iron m = 500 g initial temperature of iron t = 500 oC mass of ice M = 800 g Initial temp of the ice t ' = -10 oC Specific heat of iron c = 0.107 cal / g oC Specific heat of ice C = 0.447 cal / g oC Specific heat of water C ' = 1 cal / g oC Latent heat of fusion of water L = 80 cal / g heat lost by iron when becomes 0 oC is Q = mc dt                                                                = 500 g * 0.107 cal / g oC * 500 oC                                                                = 26750 cal heat required to come ice to 0 oC water is Q ' = MC dt '+ ML                                                                       = ( 800*0.447*10 ) + ( 800*80 )                                                                       = 3576 + 64000                                                                       = 67576 cal Q ' > Q ,So, total ice is not melted. And Q > MCdt '            >3756 cal So, the final state is ice -water mixture

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