Phenomenon: Making Liquid Nitrogen Ice Cream

Since it's been so hot out, you're in the mood for some ice cream. And, because you're a scientist, you want to use liquid nitrogen in the process. Your liquid nitrogen is stored in a Dewar (a thermally insulated container) at its boiling point. You take your liquid ice cream base from your refrigerator set at 4C (see below), put it in your kitchen aid mixer, and slowly pour liquid nitrogen into the mixing bowl. Once the liquid nitrogen touches the ice cream base, it boils.

The mixer is on the lowest setting (the energy added to the ice cream through mixing is negligible compared to the internal energy transfers). Because the nitrogen gas leaves the system it no longer exchanges energy with the ice cream base. The process happens quickly enough such that the cream and the liquid nitrogen only exchange energy with each other.

Recipe for ice "cream base":

• 2 Cups heavy cream (0.48kg)
• One 14oz can sweetened condensed milk (0.4kg)
• 2 teaspoons of vanilla extract (assumption, negligible mass compared to cream and condensed milk)

Big Question:How muchliquid nitrogen must you add to the icecream base to get the ice cream to -12C (the ideal serving temperature) while ensuring all the liquid nitrogen boils away?

Use what you've learned in PHYS 2A so far to analyze thescenario.

1. Include Temperature vs Energy and Energy Interaction diagrams as evidence in your response.Consider how many of each diagram(s) you need to fully model the process(es).
2. Use your diagrams to construct an equation that can be used toprecisely answer the big question.
3. Find a numerical answer to the Big Question or get as far as you can in that process.

Reminder: Include clear reasoning throughout your solution that justifies the choices you make.

Maybe useful information:

*Note "Cream Base" refers to the heavy cream and sweetened condensed milk mixture

TBP, Cream Base = 93C	Cp,Solid Cream Base= 1.88 kJ/(kgC)	Hp,Melting Cream Base= 210.14 kJ/(kg)
TMP, Cream Base = -2.2C	Cp Liquid Cream Base= 3.89 kJ/(kgC)
TBP, Nitrogen = -196C	Cp, Nitrogen= 1.04 kJ/(kgC)	Hp,Melting Nitrogen= 25.7 kJ/(kg)
TMP, Nitrogen = -210C		Hp,Vaporization Nitrogen= 199 kJ/(kg)
[Watt] = [Joule/s]	[kiloWatt] = [kiloJoule/s]
Ethermal = m Cp T	Ephase = m Hp
Q = E	P = Q/t or P = E/t