You buy coffee (200.0 ml) served in a styrofoam cup. You are up late working in a chemistry lab, and want to drink the coffee right away, but at 95.0 degrees C, it is too hot to drink. You like the temperature of your coffee to be 79.0 ± 1.0 degrees C. You decide that the fastest way to cool the coffee is by dropping 40.0 grams of dry ice, CO2 (s), initially at -78.0 degrees C, into the coffee. Note that CO2 does not exist as a liquid at 1 atm; the solid sublimes directly to a gas with an enthalpy of sublimation (ΔHsub) of 8.76 kJ/mol at -78.0 degrees C. Assume that the final temperature of the CO2(g) is the same as that of the coffee. You can
assume the heat capacity of coffee is the same as that for water, and you can neglect all heat losses and the solubility of CO2 in coffee. The molar mass of CO2 is 44.0 g/mol. The specific heat capacity, s = 0.850 J/goC for CO2(g);
a) Calculate the heat (qsub) required to sublime 40.0 grams of dry ice to form gaseous CO2 at -78.0 degrees C.
b) Write an expression for the quantity of heat (qgas) required to increase the temperature of the gaseous CO2 from -78.0 degrees C to the final temperature of the coffee.
c) Calculate the final temperature of the coffee in degrees C. Is the temperature of the coffee acceptable?
d) Calculate the value of ΔU for the CO2, assuming an atmospheric pressure of 1 atm.
a) To calculate the heat (qsub) required to sublime 40.0 grams of dry ice to form gaseous CO2 at -78.0 degrees C, we can use the formula:
qsub = n * ΔHsub
Where:
qsub is the heat required (in joules)
n is the number of moles of CO2 (which can be calculated using the molar mass)
ΔHsub is the enthalpy of sublimation (8.76 kJ/mol)
First, let's calculate the number of moles of CO2:
n = mass / molar mass
Given:
mass = 40.0 grams
molar mass = 44.0 g/mol
n = 40.0 g / 44.0 g/mol = 0.909 moles
Now, we can calculate the heat required to sublime the dry ice:
qsub = 0.909 moles * 8.76 kJ/mol = 7.97 kJ
Therefore, the heat required to sublime 40.0 grams of dry ice is 7.97 kJ.
b) The expression for the quantity of heat (qgas) required to increase the temperature of the gaseous CO2 from -78.0 degrees C to the final temperature of the coffee can be calculated using the formula:
qgas = m * s * ΔT
Where:
qgas is the heat required (in joules)
m is the mass of CO2 (40.0 grams)
s is the specific heat capacity of CO2 (0.850 J/goC)
ΔT is the change in temperature (final temperature of the coffee - initial temperature of CO2)
Given:
m = 40.0 grams
s = 0.850 J/goC
ΔT = final temperature of the coffee - (-78.0 degrees C) = final temperature of the coffee + 78.0 degrees C
c) To calculate the final temperature of the coffee, we need to equate the heat gained by the coffee and the heat lost by the CO2:
qcoffee = -qgas
Since we know that the heat gained by the coffee can be calculated using the formula:
qcoffee = m * c * ΔT
Where:
qcoffee is the heat gained by the coffee (in joules)
m is the mass of coffee (200.0 ml, which is equal to 200.0 grams since water has a density of 1 g/ml)
c is the specific heat capacity of water (4.18 J/goC)
ΔT is the change in temperature (final temperature of the coffee - initial temperature of the coffee)
Using the given values:
m = 200.0 grams
c = 4.18 J/goC
ΔT = (79.0 ± 1.0) degrees C - 95.0 degrees C = -16.0 ± 1.0 degrees C
Now, equating the heat gained by the coffee and the heat lost by the CO2:
m * c * ΔT = -qgas
Solving for ΔT:
ΔT = -qgas / (m * c)
Now, substitute the given values into the equation:
ΔT = -qgas / (200.0 g * 4.18 J/goC) = -qgas / 836 J
Substituting the value of qgas from part b, we get:
ΔT = -7.97 kJ / 836 J = -9.53 ± 0.01 degrees C
Finally, calculate the final temperature of the coffee:
final temperature of the coffee = initial temperature of the coffee + ΔT = 95.0 degrees C + (-9.53 ± 0.01) degrees C
Therefore, the final temperature of the coffee is 85.47 ± 0.01 degrees C.
c) The temperature of the coffee (85.47 ± 0.01 degrees C) falls within the desired range of 79.0 ± 1.0 degrees C. Therefore, the temperature of the coffee is acceptable.
d) The value of ΔU (change in internal energy) for the CO2 can be calculated using the equation:
ΔU = qsub + qgas
Substituting the values calculated in parts a and b:
ΔU = 7.97 kJ + (-7.97 kJ) = 0 kJ
Therefore, the value of ΔU for the CO2 is 0 kJ.