- #1
nooldor
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Homework Statement
The problem is this:
what is the optimum amount of dry ice inside the Styrofoam pack, if we want to received a frozen dough (temp.-18) after 3 days of transportation in temperature 20 C ?
My calculations and data placed in the attached excel file, but I'm not sure if they are correct.
Below I present my reasoning :
1. Initial temperature of the dough is -18 C deg. The dough is frozen to temp. -78 C. degrees by placing inside the Styrofoam pack with dry ice.
2. From the (Measuring the Heat of Sublimation of Dry Ice with a Polystyrene Foam Cup Calorimeter, Albert W. Burgstahler and Clark E. Bricker, 1991)
-- specific heat of dry ice is: 0.805 J/g*C
-- heat of sublimation of dry ice is: 580 J/g
and internet sources:
-- specific heat of the dough is 2.7 J/g*C
the energy equivalent in dry ice sublimation process from the second point corresponds to about 48 kg of dry ice (?, cell G11 in excel) ?
I understand that this amount sublimate and by the time of transportation of the product. It will "take" energy from product lowering its temperature.
3. When transporting in +20 C deg. outdoor temp. Heat energy will flow from the environment to the interior of the box by increasing the temperature of the product.
How much of dry ice we need to ensure product will have exactly -18 C after traveling 3 days in the Styrofoam pack in temperature +20 C deg ?
Homework Equations
q - heat flowing through the object
λ - thermal conductivity (for styrofoam: 0.036)
S - area through which the heat flows (3,66 m^2)
t - time of transportation (259200 sec)
delta T - difference of temperatures (98C = -78C to +20C)
d - styrofoam thickness (0.12m)
See excel file
The Attempt at a Solution
See excel file[/B]
