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.
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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A large punch bowl holds 3.20 kg of lemonade (which is essentially water) at 25.0 ∘C....
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