You received an ice cream maker for Christmas and bring it back with you for Spr

Question

You received an ice cream maker for Christmas and bring it back with you for Spring semester. The unit consists of a bowl and a base that rotates the bowl around a stationary mixing paddle. The walls of the bowl contain an unknown mixture that absorbs heat from the ice cream. The bowl must be frozen (-20degreeC) before use. Unfortunately, you forgot to pack the instructions and you can't remember how long you must operate the unit. No worries, you have mastered energy conservation problems (thank you BEN 201!) and calculate the amount of time required to freeze the ice cream based on the ice cream maker specifications printed on the box. Inner surface area of freezing bowl: 600 cm^2 Heat transfer coefficient for bowl: h = 0.025 J/(cm^2s-.degreeC) Power required to stir the bowl (100% efficiency): 25 W Amount of ice cream mixture added: 1 kg Ice cream mixture starts at: 10 degree C The rate of heat transfer is: Q = hA(T_bowl - T_milk) Recalling your freshman chemistry course, you remember that solutes lower the freezing point of water. Assume that the freezing point is lowered to -5 degreeC. The ice cream mixture contains milk, cream, sugar, and vanilla extract. To achieve the consistency of soft-serve ice cream, only half of the water must be frozen. At -5degree C, delta H_water = 330 kJ/kg. Derive an equation in terms of the above variables, to approximate the time required to make ice cream.

Explanation / Answer

solution:

1)here we have to freeze mixture of 1 kg of water with solute particle like sugar,milk and vanilla to it and assume they have negligible mass and they alter freezing point to -5 c

2)initially when mixture is at 10 c ,at that instant heat transfer rate is

Q'1=h*A*(Tbowl-Tmilk)

initially,Tbowl=-20 cand Tmilk=10 c,A=600 cm2,h=.025

so we get that

Q1'=-450 j/s

for instant when icecream is ready,we have

Tmilk=-5 and now we get heat transfer rate as

Q2'=h*A*(Tbowl-Tmilk)

Q2'=-225 j/s

hence average heat transfer rate over this time period is

Qm'=Q1'+Q2'/2=-337.5 j/s

3)where for mixture enthalphy has to be removed is

m=1 kg and finally alf of it has to be frozen,and hence

m1=.5 kg

dQ=m1*dH=.5*330=-165*10^3 j

where power required to stir the abll is also generate heat and added to mixture andit has to be removed with Qs'=25 j/s

4)hence energy balnce equation is,heat generated will reduce heat transfer rate

Q'=Qm'+Qs'=-337.5+25=-312.5 j/s

Q'=dQ/dt=-312.5

where dQ=-165*1063 j

hence time change from 0 sec is

dt=dQ/Q'=165*10^3/312.5=528 sec

hence approx time is 528 sec or 8.8 min and it is time of operation

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