- #1

seichan

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## Homework Statement

0.175 kg of water at 88.0 degC is poured into an insulated cup containing 0.212 kg of ice initially at 0 degC. How many kg of liquid will there be when the system reaches thermal equilibrium?

## Homework Equations

Qwater=-Qice

q=mc(Tf-Ti)

Cwater= 4187 J/kg degC

Cice=2090 J/kg degC [not sure on this one, had to look it up]

## The Attempt at a Solution

Alright, I know how to get the final temperature of the solution:

q=mc(Tf-Ti)

.175(4187)(Tf-Ti)=-.212(2090)(Tf-Ti)

.175(4187)Tf-.175(4220)(88+273)=-.212(2090)Tf

.175(4187)Tf+.212(2090)Tf=.175(4187)(88)

Tf=[.175(4187)(88)]/[.175(4187)+.212(2090)]

What I'm not sure of is how much of the ice this melts into water... I considered putting the temperature back into the equilibrium equation and solving for how much mass it must take, but I'm very confused as to how to denote the change in mass, considering the fact that the final masses of both the ice and the water are unknown.

Qwater=-Qice

(mi(water)-mf(water))(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)]-88)=-(mi(ice)-mf(ice)(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)])

(.175-mf(water))(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)]-88)=-(.212-mf(ice)(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)])

Any help would be appreciated.

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