What Is the Final Temperature When Ice Melts in Water?

[cugirl]

Homework Statement

An insulated cup contains 1kg of water initially at 20 C. 0.50 kg of ice, initially at
0 C is added to the cup of water. The water and ice are allowed to come to thermal equilibrium. The specific heat of ice is 2000 J/kg oC, the specific heat of water 4186 J/kg oC, the latent heat of fusion of water is 33.5x104J/kg. What is the final temperature of the water?

(A) 0 C
(B) 1.2 C
(C) 4.6 C
(D) 9.2 C
(E) 15.2 C

Homework Equations

Q_gain(ice) = Q_lost(water)
mc(Ti-Tf) = mLf + mc(Tf-0)

The Attempt at a Solution

Q_gain(ice) = Q_lost(water)
mc(Ti-Tf) = mLf + mc(Tf-0)
(1.0)(4186)(20 - Tf) = (.5)( 33.5x104 ) + (.5)(4186)(Tf)
83,720 – 4186Tf = 167,500 + 2093 Tf
-83780= 6279 Tf
-13.3 = Tf

 

[mgb_phys]

Correct method
But what is the total energy needed to melt the ice?
How much energy can you get from the initial water before it cools to 0c?

 

[nlightner]

The final temperature of the water is (A) 0 C.


I would like to commend you for using the correct equations and units in your attempt at solving this problem. However, I would like to point out that your final answer of 0 degrees Celsius does not make sense physically. This would mean that the ice did not melt at all and the water did not change temperature, which is not possible.

To find the correct answer, we need to set the two equations equal to each other and solve for Tf. This will give us the temperature at which the water and ice are in thermal equilibrium.

Q_gain(ice) = Q_lost(water)
mc(Ti-Tf) = mLf + mc(Tf-0)
(1.0)(2000)(0 - Tf) = (.5)( 33.5x10^4 ) + (.5)(4186)(Tf)
-2000Tf = 167500 + 2093Tf
-4093Tf = 167500
Tf = 40.9 degrees Celsius

Therefore, the final temperature of the water is (E) 40.9 degrees Celsius. This makes more sense physically as the ice would have melted and the water would have cooled to reach thermal equilibrium.