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Heat and final temperature problem

  • Thread starter narutoish
  • Start date
25
0
1. Homework Statement

The "steam" above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature of 290 g of hot coffee initially at 90.0°C if 1.50 g evaporates from it? The coffee is in a Styrofoam cup, and so other methods of heat transfer can be neglected. Assume that coffee has the same physical properties as water.

2. Homework Equations

Q = mcΔT thats the only equation that i saw fit, but i dont know how to utilize it in this situation

3. The Attempt at a Solution

the answer is 87.2

i tried writing two heat equations like (290g)(Tf- 900C) = (1.50)(100)
but i don't know if thats right. please point me in the right direction

thanks
 

CAF123

Gold Member
2,888
88
Some of the water is evaporated as a result of adding energy to the cup+water system. So you can think of energy Q = cmΔT entering the system, and the mass of water that evaporates carries Q' out of the system. Find what this Q' is and express the above mathematically.
 
19,458
3,893
What is the heat of vaporization of water in cal/gm? What is the heat capacity of liquid water in cal/(gm-C)? How much heat is required to evaporate 1.5 gm of water?

Chet
 
25
0
So I set the heat equation up like this Q = (1.50g)(4.184)(-10 0C ) and find Q to be -62.76.

Then I set that Q equal to the other heat equation like : -62.76 = (288.5g)(4.184)(Tf-90) but that gives me 89.9 , am I going in the right direction?
 
19,458
3,893
So I set the heat equation up like this Q = (1.50g)(4.184)(-10 0C ) and find Q to be -62.76.

Then I set that Q equal to the other heat equation like : -62.76 = (288.5g)(4.184)(Tf-90) but that gives me 89.9 , am I going in the right direction?
No. Your first equation is wrong. You should be multiplying the 1.5 g by the heat of vaporization of water.

Chet
 
25
0
Thanks I got it now. But is reason I couldn't use the Q =mcT equation because of the phase change?

Thanks Chet
 

CAF123

Gold Member
2,888
88
Hi Chestermiller,

I was just wondering about the physical reasoning for setting those two equations equal to each other. If we consider the cup+water as the system, then energy Q1(=cwmw(90-To) enters the system to get it to the temperature of 90 degrees C. The energy of the cup+water system is then conserved if there are no other heat transfers or work done upon it. So by conservation of energy, we have Q1 = cwm1(Tf-90) + m2Lv, where cw = specific heat of water, m1 = mass of water left in cup, m2 = mass evaporated and Lv = latent heat of vaporization.

The differential (wrt time) of the LHS is zero, so the differential of the RHS must be zero. How would I show from this that cwm1(Tf-90) = - m2Lv follows?

Thanks.
 
19,458
3,893
Hi Chestermiller,

I was just wondering about the physical reasoning for setting those two equations equal to each other. If we consider the cup+water as the system, then energy Q1(=cwmw(90-To) enters the system to get it to the temperature of 90 degrees C. The energy of the cup+water system is then conserved if there are no other heat transfers or work done upon it. So by conservation of energy, we have Q1 = cwm1(Tf-90) + m2Lv, where cw = specific heat of water, m1 = mass of water left in cup, m2 = mass evaporated and Lv = latent heat of vaporization.

The differential (wrt time) of the LHS is zero, so the differential of the RHS must be zero. How would I show from this that cwm1(Tf-90) = - m2Lv follows?

Thanks.
Hi CAF123. I think my understanding for the problem statement is different from yours. The problem statement says that the coffee starts off at 90C, and that the coffee cup is adiabatic: "The coffee is in a Styrofoam cup, and so other methods of heat transfer can be neglected."

So, what I get out of this is that Q = 0. So, cwm1(Tf-90) + m2Lv =0.

Hope this makes some sense.

Chet
 

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