Question

A large punch bowl holds 4.00 kg of lemonade (which is essentially water) at 24.0 ∘C. A 6.00×10⁻²-kg ice cube at -12.0 ∘C is placed in the lemonade.

1.What is the final temperature of the system? Ignore any heat exchange with the bowl or the surroundings.

2.What is the amount of ice (if any) remaining?

Answer 1

Given
    mass of lemonade is M = 4.00 kg

    mass of ice is m = 6.00*10^-2 kg
temperature of the lemonade is Tl = 24 0C
temperature of the ice is Ti = -12 0C
the specific heat of water is Cw = 4181.3 J/kg.k
specific heat of ice is Ci = 2110 J/kg.k
latent heat of fusion of ice is L = 3.33*10^5 J/kg
when the ice is placed in lemonade , ice gains heat energy from water and water loose heat energy , so that the ice
so that if we consider the final temperature of the system is 0 0C where the ice exists then we can say certain amount of ice will remain so
heat lost by water is  
    Q = M*C*dT
    Qw = 4*4181.3*24 J

    Qw = 401404.8 J


and that of ice gain heat energy is  
    Qi = 6.00*10^-2*2110*12 + 6*10^-2*3.33*10^5 J
    Qi = 21500J

QW > Qi


all the ice melts




means the ice will completely melt so  
the equation can can be written as
heat lost by water = heat gain by ice

    Mw*Cw*(T-Tw) = mi*Ci(0-Ti)+mi*L + mi*Cw*(T-0)

    4*4181.3*(24-T) = (6.00*10^-2*2110*12)+(6.00*10^-2*3.33*10^5) + (6*10^-2*4181.3*T)
    T = 22.4 0C
final temperature is T = 22.4 0C