# Heat and Energy Transfer Help

## Homework Statement

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?

Q1=Q2
Q=mCpΔT
Q=mLvap
Q=mLfus

## 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$\cdot$C°
CpAg = 230 J/kg$\cdot$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!!

Spinnor
Gold Member
Look at the largest term on the right, it dominates,

(0.001)(4186)(100) = 418

and make the left side as small as possible,

x(130)(964) + (0.02-x)(230)(862) --> .02*130*964 = 2500

The problem as stated is wrong or your values are wrong?

@Spinnor --thank you! I figured out what was wrong --both my values for heat of fusion and heat of vaporization were wrong --I had to multiply them each by 103 to get the right answer, whoops! Thanks for the help though!