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## 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