Question

A large punch bowl holds 3.20 kg of lemonade (which is essentially water) at 25.0 ∘C. A 5.40×10−2-kg ice cube at -12.0 ∘C is placed in the lemonade. What is the final temperature of the system? Ignore any heat exchange with the bowl or the surroundings.

Answer 1

Suppose the final temperature of system is T.

Now Using energy conservation:

Heat gained by ice cube = Heat released by lemonade

Q1 = Q2

Q1 = Heat gained by ice from -12 C to 0 C + heat gained during phase change + heat released from 0 C to T

Q1 = Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT

Q2 = Heat released by lemonade = Mw*Cw*dT2

Given Values are:

Mi = Mass of ice = 5.40*10^-2 kg

Mw = Mass of lemonade = 3.20 kg

Ci = Specific heat of ice = 2090 J/kg-C

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

Lf = Latent heat of fusion of water = 3.34*10^5 J/kg

dT1 = 0 - (-12) = 12 C

dT = T - 0 = T

dT2 = 25 - T

So Using these values:

Q1 = Q2

5.40*10^-2*2090*12 + 5.40*10^-2*3.34*10^5 + 5.40*10^-2*4186*T = 3.20*4186*(25 - T)

T = [3.20*4186*25 - (5.40*10^-2*2090*12 + 5.40*10^-2*3.34*10^5)]/(5.40*10^-2*4186 + 3.20*4186)

T = 23.2 C = Final temperature of lemonade

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