Molten iron at 1538 degree C (its melting point) is solidified by an ice-water m

Question

Molten iron at 1538 degree C (its melting point) is solidified by an ice-water mixture at 0 degree C. Assuming that there is still ice present when the process is over and the iron is left in the ice-water, how much ice will be melted if 50.0 g of iron is solidified? The following may be useful: enthalpy of fusion of iron = 13.8 kJ/mol; enthalpy of fusion of water = 6.07 kJ/mol; specific heat capacity of iron = 0.449 J/g-K; specific heat capacity of water = 4.184 J/g-K. 102 g 36.6 g 87.8 g 139 g 46.9 g

Explanation / Answer

While solidifying iron it loses = (50.0/ 56) * 13.8 *1000 = - 12321.43 J

6070 J of heat can melt 18 g. of ice

then, 12321.43 J heat can melt 18 * 12321.43 / 6070 = 36.6 g.

So, the answer is (2) 36.6 g.

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