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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.
 
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3,840
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?
 
19,357
3,840
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.
 
19,357
3,840
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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