What is the amount of steam needed to heat 0.50kg of diet cola to 90 deg C?

[03cobra 87gn]

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

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

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 don't know which or if either one are right so any help would be greatly appreciated
thanks

 

[Shooting Star]

Your eqn is not correct.

The temp of the water rises from 20 deg (not mentioned in problem, but used in eqn) to 90 deg. This will be your LHS. The steam is presumably at 100 deg. The steam gives off latent heat and then cools down to 90 deg. This will be your RHS.

Now set up the eqn. If you are using a consistent system of units. like the CGS, you don't have to write the units in the middle of the eqns -- only the values will do.

 

[03cobra 87gn]

20 deg C was not listed in the prob. so i used it i remember our teacher saying if its not listed in the prob. to use 20 deg C.

so my eqn would look like this? 500 * 1 (20-90) + M steam * .480 (100-90)

M steam = 7.3 kg

i don't have the answer to the prob. to check to see if its right.

 

[Shooting Star]

so my eqn would look like this? 500 * 1 (20-90) + M steam * .480 (100-90)

That's not an eqn -- where's the equality sign? And what is this 0.480? In CGS, the latent heat of evaporation is 540 cal/g.

If M is the mass of steam reqd, your RHS should have M*540+M*1*(100-90).

The water gains heat, and its temp rises from 20 C to 90 C. This heat is supplied by the steam condensing into water and giving up the latent heat, and then this water at 100 C cooling down to 90 C.

Form the eqn and solve it. (It's already almost done.)

 

[03cobra 87gn]

I was doing it differently I was using the conservation of energy SDQ=0

So I was doing DQcola + DQsteam =0 (cola in prob. said to be mostly water so I used water)

I did M cola * c cola * DT + M steam * L vap. = 0

500 * 1 * (20-90) + M steam * 540 =0

540 M steam = 35000

M steam = 68.4g and all the units cancel out nicely and I’m just left with grams

Thanks for the help!

 

[Shooting Star]

So I was doing DQcola + DQsteam =0 (cola in prob. said to be mostly water so I used water)

I did M cola * c cola * DT + M steam * L vap. = 0

500 * 1 * (20-90) + M steam * 540 =0

540 M steam = 35000

M steam = 68.4g and all the units cancel out nicely and I’m just left with grams

You did not take into consideration the decrease in the temp of the water formed from steam. How can your answer be correct?

 

[03cobra 87gn]

well i the question stated heating .50 kg of cola (mostly water), by adding steam. How musch steam should be bubbled in so that the final temp. is 90 deg C.

isnt the steam just being used as the heat source? so your saying that i have to take into account the steam that turns back into water when it hits the cola? the question never said if or what the cola is in. where would i put the other change in temp at?

 

[Shooting Star]

isnt the steam just being used as the heat source? so your saying that i have to take into account the steam that turns back into water when it hits the cola? the question never said if or what the cola is in. where would i put the other change in temp at?

Now you discussing good Science. The steam has to condense to water at 100 C, and this water will cool down, and it will be in the mixture. The final result has to take that into account.

Heat gained by the colder objects = heat lost by the hotter objects.

Read post #4 again and carefully this time.

 

[03cobra 87gn]

So $$\Delta$$Qsteam +$$\Delta$$Qcola=0

Heat lost by one system = heat gaind by the other

DQ steam = DQ cola

In my book the formula latent heat is Q=ML there is no change in temp

500 * 1 * (20-90) = M steam 540

Where would I put change in temp of the steam from 100-90?? The Lv for steam doesn’t have deg C in it.


OR is this what you mean

so my RHS would be M steam * 540 + M water condensed * 1 * (100-90)?
and the LHS would be M cola *1*(20-90)

M cola *1*(20-90) = M steam * 540 + M water condensed * 1 * (100-90) i don't know the other mass, or am i still just not getting

 

[Shooting Star]

OR is this what you mean

so my RHS would be M steam * 540 + M water condensed * 1 * (100-90)?
and the LHS would be M cola *1*(20-90)

M cola *1*(20-90) = M steam * 540 + M water condensed * 1 * (100-90) i don't know the other mass, or am i still just not getting

Correct 20-90 to 90-20 above.

You are there, but you don't know it. Give it a few minutes.

If some steam condenses into water, does the MASS change?

That's it, mark it solved after you actually solve it. :zzz:

 

[03cobra 87gn]

oohhh i totally missed that i got it now. Thanks for all the help i might be posting another one that's giving me trouble, I'm gona see if i can figure it out.