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**1. Homework Statement**

heating .50kg of diet cola mostly water, by

adding steam. How much steam should be bubbled in so final temp. is 90 deg C

**2. Homework Equations**what i tired so far downt know if it the right way to go abotu the problem is specific heat capacity Q=mc detlaT

E tot of the system =0 so Q1+Q2 =0

**3. The Attempt at a Solution**

so i did Q(cola)+Q(steam)=0

M(cola)*C(cola)*(Tf-Ti)+M(steam)*C(steam)*deltaT

500g*1 cal/g*deg C (90-20 deg C) + M steam * .480 cal/g*deg C * delta T

the last delta T on the right do i keep that in or do i use the temp. of steam in my book?

if i use 7 deg C for the stema delta T i get 1.04 kg of steam

if i use the 110 deg C which is under the specific heat of steam i get .662 kg

i dont know which or if either one are right so any help would be greatly appreciated

thanks