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## 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

Q

_{1}=Q

_{2}

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[itex]\cdot[/itex]C°

Cp

_{Ag}= 230 J/kg[itex]\cdot[/itex]C°

m

_{Au}Cp

_{Au}ΔT

_{Au}+ m

_{Ag}Cp

_{Ag}ΔT

_{Ag}= m

_{ice}Cp

_{ice}ΔT

_{ice}+ m

_{ice}L

_{fus}+ m

_{water}Cp

_{water}ΔT

_{water}+ m

_{water}L

_{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!!