Posted by Jen on .
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.

Chemistry/Calorimetry 
Anonymous,
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