Heat and final temperature problem

In summary, the conversation discusses finding the final temperature of a cup of hot coffee after some of the water has evaporated. The solution involves using the conservation of energy equation, setting the energy entering the system equal to the energy leaving the system, and solving for the final temperature. It is also noted that the heat of vaporization of water must be taken into account in the calculations.
  • #1
narutoish
25
0

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.

Homework Equations



Q = mcΔT that's the only equation that i saw fit, but i don't know how to utilize it in this situation

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 that's right. please point me in the right direction

thanks
 
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  • #2
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.
 
  • #3
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
 
  • #4
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?
 
  • #5
narutoish said:
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
 
  • #6
Thanks I got it now. But is reason I couldn't use the Q =mcT equation because of the phase change?

Thanks Chet
 
  • #7
narutoish said:
Thanks I got it now. But is reason I couldn't use the Q =mcT equation because of the phase change?
Yes.
 
  • #8
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.
 
  • #9
CAF123 said:
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
 

1. What is the heat and final temperature problem?

The heat and final temperature problem is a common physics problem that involves calculating the final temperature of a system after heat is transferred between two or more objects with different initial temperatures.

2. What factors affect the final temperature in the heat and final temperature problem?

The final temperature in the heat and final temperature problem is affected by the initial temperatures of the objects, the amount of heat transferred, and the specific heat capacities of the objects.

3. How do you solve the heat and final temperature problem?

The heat and final temperature problem can be solved using the equation Q = mcΔT, where Q is the heat transferred, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.

4. What is the difference between specific heat capacity and heat capacity?

The specific heat capacity is the amount of heat required to raise the temperature of one unit of mass by one degree, while the heat capacity is the amount of heat required to raise the temperature of an entire object by one degree.

5. Can the heat and final temperature problem be applied to real-world situations?

Yes, the heat and final temperature problem can be applied to real-world situations such as cooking, heating and cooling systems, and thermal energy storage.

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