Ice Cream Lab
Ice Cream Lab
Purpose: To use the properties of thermodynamics to create delicious vanilla ice cream.
Recipe:
2 cups heavy cream
1 cup whole milk
2/3 cup sugar
1 teaspoon vanilla extract
1 vanilla bean, scraped
Data Table:
Time mixed: 30 minutes, 1800s Mass of salt: 1.77 kg
Mass of ice: 2.27 kg
Initial Temp. (ingredients): 12 Degrees Celsius Initial Temp. (ice) -12 Degrees Celsius
Total mass: 0.64 kg Final temp. (ice and ice cream): 4 Degrees Celsius
Mass of ingredients before: 1.13 kg Voltage: 120 V
Current: 1.2 A
Work done by ice cream maker: P=VI P=(120 V)(1.2 A) P= 144W
P=W/t W=(P)(t) W= (144 W)(1740 s) W= 250560 J
Heat Gained by ice (as it cools and melts): q=mcΔt = (4.082 kg)(4186 J/kg°C)(13°C)
= 222,134.276 J q=mL = (1.25kg)(3.34x105) = 417,500 J
222,134.276 J + 417,500 J = 639,634.276 J
Heat lost by metal bucket: q=mcΔt = (0.196 kg)(897 J/kg°C)(23.3°C-5°C) = 3217.36 J
Heat lost by ice cream: QIce Cream= Q1+Q2 Q1=mcΔt Q2=mL
QIce Cream = (0.666kg)(3000J/kg°C)(13°C)+(0.666kg)(2.1014x105 J/kg)
QIce Cream =165,869.938 J
Total internal energy change for ice cream: E= -QIce - QBucket – QIce Cream – W
E= -250,560 J – 639,634.276 J – 3217.36 J – 165,869.938 J
E= -1,059,281.574
Work of Ice Cream Maker:
P=VI P=(120 V)(1.2 A)
P=W/t W=(P)(t) W= (144 W)(1800 s) W= 259kJ
Heat gained by ice: q=mcΔt = (4.082 kg)(4186 J/kg°C)(13°C)
= 222,134.276 J q=mL = (1.25kg)(3.34x105) = 417,500 J
222,134.276 J + 417,500 J = 639,634.276 J
Heat lost by metal bucket: q=mcΔt = (0.196 kg)(897 J/kg°C)(23.3°C-5°C) = 3217.36 J
Heat lost by ice cream:
QIce Cream= Q1+Q2 Q1=mcΔt Q2=mL
QIce Cream = (0.666kg)(3000J/kg°C)(13°C)+(0.666kg)(2.1014x105 J/kg)
QIce Cream =165,869.938 J
Ice cream energy change:
E= -QIce - QBucket – QIce Cream – W
E= -250,560 J – 639,634.276 J – 3217.36 J – 165,869.938 J
E= -1,059,281.574
Analysis: 1) Was the change in internal energy negative
Purpose: To use the properties of thermodynamics to create delicious vanilla ice cream.
Recipe:
2 cups heavy cream
1 cup whole milk
2/3 cup sugar
1 teaspoon vanilla extract
1 vanilla bean, scraped
Data Table:
Time mixed: 30 minutes, 1800s Mass of salt: 1.77 kg
Mass of ice: 2.27 kg
Initial Temp. (ingredients): 12 Degrees Celsius Initial Temp. (ice) -12 Degrees Celsius
Total mass: 0.64 kg Final temp. (ice and ice cream): 4 Degrees Celsius
Mass of ingredients before: 1.13 kg Voltage: 120 V
Current: 1.2 A
Work done by ice cream maker: P=VI P=(120 V)(1.2 A) P= 144W
P=W/t W=(P)(t) W= (144 W)(1740 s) W= 250560 J
Heat Gained by ice (as it cools and melts): q=mcΔt = (4.082 kg)(4186 J/kg°C)(13°C)
= 222,134.276 J q=mL = (1.25kg)(3.34x105) = 417,500 J
222,134.276 J + 417,500 J = 639,634.276 J
Heat lost by metal bucket: q=mcΔt = (0.196 kg)(897 J/kg°C)(23.3°C-5°C) = 3217.36 J
Heat lost by ice cream: QIce Cream= Q1+Q2 Q1=mcΔt Q2=mL
QIce Cream = (0.666kg)(3000J/kg°C)(13°C)+(0.666kg)(2.1014x105 J/kg)
QIce Cream =165,869.938 J
Total internal energy change for ice cream: E= -QIce - QBucket – QIce Cream – W
E= -250,560 J – 639,634.276 J – 3217.36 J – 165,869.938 J
E= -1,059,281.574
Work of Ice Cream Maker:
P=VI P=(120 V)(1.2 A)
P=W/t W=(P)(t) W= (144 W)(1800 s) W= 259kJ
Heat gained by ice: q=mcΔt = (4.082 kg)(4186 J/kg°C)(13°C)
= 222,134.276 J q=mL = (1.25kg)(3.34x105) = 417,500 J
222,134.276 J + 417,500 J = 639,634.276 J
Heat lost by metal bucket: q=mcΔt = (0.196 kg)(897 J/kg°C)(23.3°C-5°C) = 3217.36 J
Heat lost by ice cream:
QIce Cream= Q1+Q2 Q1=mcΔt Q2=mL
QIce Cream = (0.666kg)(3000J/kg°C)(13°C)+(0.666kg)(2.1014x105 J/kg)
QIce Cream =165,869.938 J
Ice cream energy change:
E= -QIce - QBucket – QIce Cream – W
E= -250,560 J – 639,634.276 J – 3217.36 J – 165,869.938 J
E= -1,059,281.574
Analysis: 1) Was the change in internal energy negative