Thermodynamics- How much Ice is melted?

[seichan]

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.

 

[alphysicist]

Hi seichan,

You do use $q=m c (\Delta T)$ for the heat flow in or out required to change the temperature of a mass m of a substance. However, here the ice is initially at zero degrees celsius. As the heat initially begins flowing into the ice, it begins melting, but it's temperature does not change until it completely melts. How is heat flow related to the mass of ice that has melted?

 

[Moetasim]

I would approach this problem by first understanding the concept of thermal equilibrium. When two objects at different temperatures come into contact, they will exchange heat until they reach the same temperature. In this case, the water at 88.0 degC will transfer heat to the ice at 0 degC until they both reach the same temperature.

Using the equation Qwater = -Qice, we know that the amount of heat lost by the water (Qwater) will be equal to the amount of heat gained by the ice (Qice). We can calculate Qwater and Qice using the specific heat capacity (c) and the change in temperature (ΔT).

Qwater = mcΔT = (0.175 kg)(4187 J/kg degC)(88.0 degC - Tf)
Qice = mcΔT = (0.212 kg)(2090 J/kg degC)(Tf - 0 degC)

Since Qwater = -Qice, we can set these two equations equal to each other and solve for Tf:

(0.175 kg)(4187 J/kg degC)(88.0 degC - Tf) = -(0.212 kg)(2090 J/kg degC)(Tf - 0 degC)
Solving for Tf, we get Tf = 0.67 degC. This means that the final temperature of the system will be 0.67 degC.

Now, to determine the amount of ice that has melted into liquid water, we can use the equation Qice = mL, where L is the latent heat of fusion (amount of heat required to melt 1 kg of ice). The latent heat of fusion for water is 334,000 J/kg.

Qice = mL = (0.212 kg)(334,000 J/kg) = 70,808 J

We know that Qice = -Qwater, so the amount of heat lost by the ice (70,808 J) will be equal to the amount of heat gained by the water. We can use the equation Qwater = mcΔT to solve for the mass of water (m) that has gained this amount of heat:

Qwater = mcΔT = (m)(4187 J/kg degC)(0.67 degC - 88.0 degC)
Solving for m, we get m = 0.017 kg. This means that