Twenty grams of silver and gold (total mass of the gold plus the silver is 20 grams) is used to convert 1 g of ice at -3°C to steam at 100°C. If both the silver and the gold are at their melting points at the beginning and the final temperature of the silver, gold and steam is 100°C, what was the individual mass of the silver and the gold?
The Attempt at a Solution
mAu= x kg
mAg= (.02-x) kg
mice= .001 kg
ΔTAu= 1064-100 = 964
ΔTAg= 962-100 = 862
CpAu = 130 J/kg[itex]\cdot[/itex]C°
CpAg = 230 J/kg[itex]\cdot[/itex]C°
mAuCpAuΔTAu + mAgCpAgΔTAg = miceCpiceΔTice + miceLfus + mwaterCpwaterΔTwater + mwaterLvap
x(130)(964) + (0.02-x)(230)(862) = (0.001)(2100)(3) + (0.001)(333) + (0.001)(4186)(100) + (0.001)(2260)
125320x + 3965.2 - 198260x = 6.3 + 0.333 + 418.6 + 2.26
-72940x = -1980.5
x = 0.027 kg
My answer is over 0.02 kg (20g) and because I'm solving for the mass of gold, it was to be less than 20 grams. I've done this problem about four times already and I know I'm doing something wrong, I just can't figure out what.
Any help is greatly appreciated, thank you!!