What Are the Individual Masses of Silver and Gold Used to Convert Ice to Steam?

[horcruxes]

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?

Homework Equations

Q₁=Q₂
Q=mCpΔT
Q=mL_vap
Q=mL_fus

The Attempt at a Solution

m_Au= x kg
m_Ag= (.02-x) kg
m_ice= .001 kg
ΔT_Au= 1064-100 = 964
ΔT_Ag= 962-100 = 862
Cp_Au = 130 J/kg\cdotC°
Cp_Ag = 230 J/kg\cdotC°m_AuCp_AuΔT_Au + m_AgCp_AgΔT_Ag = m_iceCp_iceΔT_ice + m_iceL_fus + m_waterCp_waterΔT_water + m_waterL_vap
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]

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?

 

[harrysawizard]

@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!