Time required for a liquid to change in Temp (thermo)

[movetwice]

Hi this is my first post here.

I am attempting to calculate how long it takes for a soda can(355ml) to change from Current room temperature(75F), to say 38F, however the refrigerator or ambient temperature in this case is 35F.

I was given a corollary from a friend, but the values I get in return are way off.

t = pV/hA * LN[ (T1 - Ta)/(T2 - Ta) ]

t= [(988.9 kg/m3 )(0.000355 m3 )]/[(255W/m2K)(.0277m2)] * LN[ (75 - 35)/(38 - 35)]

I attempted to use metric for all the units.
p=density
V=Volume
A=surface area
h= convection coefficient of water or the aluminum can?
T1 and T2 are the initial and final temperatures of the drink.
Ta is the ambient temperature.

However I plug-in the equation it seems to not output realistic data. I'll get something like .5 seconds depending on what variables I plug in/manipulate.

Is my equation incorrect or my units incorrect? Also are their any other better equations to do this with? This is driving me mad, please help.

Ideally I need to figure out a working base equation, as well as know what each unit is.

 

[Mapes]

Hi movetwice, welcome to PF. Your equation has to be wrong, because it gives time in units of kg W K^-1!

Presumably your friend is assuming the temperature of the liquid in the can is uniform, and the heat lost equals the heat transferred by convection:

$$\rho Vc\frac{dT}{dt}=hA(T-T_\infty)$$

which is solved to give

$$t=\frac{\rho Vc}{hA}\ln\frac{T_i-T_\infty}{T-T_\infty}$$

And the convection coefficient looks way too high. I would think 25-50 W m^-2 K^-1 would be more reasonable, but others may want to weigh in with their estimates.

 

[movetwice]

excellent thank you!, this might be a dumb follow up question, but what is the unit for "c"?

 

[Mapes]

$c$ is the http://en.wikipedia.org/wiki/Specific_heat" .

 

[movetwice]

thanks!

 

[movetwice]

Alright so I'm still running into issues with this equation. I'm thinking my units are off somehow, because allegedly a can of soda is supposed to get to optimal temperature in 20-25 minutes in a freezer.

t = pVc/hA * LN[ (T1 - Ta)/(T2 - Ta) ]
t= [(988.9 kg/m3 )(0.000355 m3 )(4.186)]/[(25W/m2K)(.000277m2)] * LN[ (75 - 0)/(38 - 0)]

t= 2.12207 * 0.679901 = 1.44279

p = 998.9 kg/m3 (standard density of water)
V= 355ml = .000355 m3
C= standard C of water 4.186 J/g*C
h= 25 W/m2k
A= 277 cm^2 = .0277 m^2 (or sqm)

Any advice on where I am going wrong? I'm looking to get a working equation that will be highly accurate so I can isolate and change certain variables and get real usable time calculations.

 

[Mapes]

Have you checked your units? At the end, you should be left with seconds.

 

[GT1]

H for free convection is usually not more then 5-7 W/m^2*K
Also the temperature should be in K or degC

 

[nasu]

Alright so I'm still running into issues with this equation. I'm thinking my units are off somehow, because allegedly a can of soda is supposed to get to optimal temperature in 20-25 minutes in a freezer.

t = pVc/hA * LN[ (T1 - Ta)/(T2 - Ta) ]
t= [(988.9 kg/m3 )(0.000355 m3 )(4.186)]/[(25W/m2K)(.000277m2)] * LN[ (75 - 0)/(38 - 0)]

t= 2.12207 * 0.679901 = 1.44279

p = 998.9 kg/m3 (standard density of water)
V= 355ml = .000355 m3
C= standard C of water 4.186 J/g*C
h= 25 W/m2k
A= 277 cm^2 = .0277 m^2 (or sqm)

Any advice on where I am going wrong? I'm looking to get a working equation that will be highly accurate so I can isolate and change certain variables and get real usable time calculations.

For consistent units, you need c in J/kg*C, which is about 4190.
This will change the result by a factor of 1000.