A 50.0 g piece of iron at 156 C is dropped into 20.0g H2O(l) at 90C in an open,

Question

A 50.0 g piece of iron at 156 C is dropped into 20.0g H2O(l) at 90C in an open, thermally insulated container. How much water would you expect to vaporize, assuming no water splashes out? The specific heats of iron and water are 0.45 and 4.21 J / G*C respectively, and Delta H{vap}=40.7 kJ/ mole H2O. (Answer in grams)

Explanation / Answer

heat lost by iron = 50*0.45*(156-100) Let 'x' grams of water vaporizes => x/18 moles of water vaporizes heat gained by water = 20(4.21)(100-96)+ (x/18)*(40.7*10^3) SInce the container is thermally insulated, hence, heat gained by water = heat lost by iron => 50*0.45*(156-100) = 20(4.21)(100-96)+ (x/18)*(40.7*10^3) => (x/18)*(40.7*10^3) = 418 => x = 0.185 grams => 0.185 grams of water vaporizes

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