Question

A large punch bowl holds 3.30 kg of lemonade (which is essentially water) at 22.0 ∘C. A 5.60×10−2-kg ice cube at -10.0 ∘C is placed in the lemonade.

Part A What is the final temperature of the system? Ignore any heat exchange with the bowl or the surroundings. T T = nothing   ∘C   SubmitRequest Answer Part B What is the amount of ice (if any) remaining? Express your answer using one significant figure. m m = nothing   kg   SubmitRequest Answer Provide Feedback

Answer 1

Part A.

Assuming all the ice is melt and final temperature of system is T, then if final temperature comes negative, than some there must be some amount of ice remaining, And if final answer is positive, than that temperature is final temperature of system and mass of ice will be 0 gm

Suppose the final temperature is T.

Now Using energy conservation:

Heat gained by ice = Heat released by lemonade

Q1 + Q2 + Q3 = Q4

mi*Ci*dT1 + mi*Lf + mi*Cw*dT2 = m1*Cw*dT3

Q1 = Heat absorbed by ice from -10 C to T C

Q2 = Heat absorbed during phase change

Q3 = Heat absorbed by ice from 0 C to T

dT1 = 0 - (-10) = 10

dT2 = T - 0 = T

mi = 5.60*10^-2 kg,

Ci = 2090 J/kg-C

m1 = mass of lemonade = 3.30 kg

Cw = Specific heat capacity of water = 4186 J/kg-C

dT3 = 22.0 - T

Lf = 3.33*10^5 J/kg = latent heat of fusion

Now using given values:

5.60*10^-2*2090*10 + 5.60*10^-2*3.34*10^5 + 5.60*10^-2*4186*T = 3.30*4186*(22 - T)

Now Solving above equation

T = 20.2 C

Since T > 0, So final amount of ice = 0 gm

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