Temperature Fusion (heat gain = heat loss)

In summary, according to the student, the equation to solve for the temperature of the resulting mixture is (10+2)*540 +2*100-10*80 = 480. The student arrived at the correct answer of 40 degrees Celsius by solving for the temperature using the heat evolve=heat absorb equation and using the temperature of steam at 100 degrees Celsius.
  • #1
aodhowain
4
0

Homework Statement


If 10 kilogram of ice at zero degrees celsius is added to 20 kilogram of steam at 100 degrees celsius, what is the temperature of the resulting mixture?

Homework Equations


uhm... perhaps how many solutions are required to arrive at the answer?

The Attempt at a Solution


I and some of my classmates arrived at 16.11 degrees celsius but my instructor said its wrong. He said the answer is around 40 degrees celsius. Now, he gave it as an assignment to us to find the correct solution.
 
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  • #2
In no way that's advanced physics.

Show how you got 16 deg answer. What equations have you used?
 
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  • #3
Borek said:
What equations have you used?

QE = QA
Heat Evolve = Heat Absorb
MECEdelta TE = MACAdelta TA
 
  • #4
aodhowain said:
QE = QA
Heat Evolve = Heat Absorb

So far so good.

MECEdelta TE = MACAdelta TA

That's not all.

I have asked you to show how you got the wrong answer.
 
  • #5
MECEdelta TE = MACAdelta TA
(10 kg)(0.50 cal/kg*C)(tf - 0C) = (2 kg)(0.48 cal/kg*C)(100C - tf)
5 cal/C * tf = 96 cal - 0.96 cal/C * tf
5.96 cal/C * tf = 96 cal
tf = (96 cal)/(0.96 cal/C)
tf = 16.11 C
 
  • #6
As I told you - you are missing part of the equation. Two parts, to be precise. What you did would be OK for a mixture of water at 0 deg C and water at 100 deg C - but you start with ice and steam.
 
  • #7
Borek said:
As I told you - you are missing part of the equation. Two parts, to be precise. What you did would be OK for a mixture of water at 0 deg C and water at 100 deg C - but you start with ice and steam.

I agree. A bit more help, heat from hot water has to be transferred to the ice to melt it, even though the temperature does not change. Quantify this heat.
 
  • #8
I think I got the solution to arrive at 40 degrees celsius. Please verify if my answer is correct.

Q out of steam = Q into ice (for c I use kcal/kgC but You can just as easily use J/kgC)

Q out = m*Lv + m*c*deltaT = m*(Lv + c*deltaT) = 2.0kg*(540kcal/kg + 1.0kcal/kg/C*(100-Tf))

Q in = m*Lf +m*c*deltaT = m*(Lf =c*deltaT) = 10kg*(80kcal/kg + 1.0kcal/kgC*(Tf -0))

Setting these equal gives 2.0*540 +2.0*100 -2.0*Tf = 10*80 + 10*Tf

Solving (10+2)*Tf = 2*540 +2*100-10*80 = 480 so Tf = 480/12 = 40degC
 
  • #9
I have just skimmed, looks reasonably.

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FAQ: Temperature Fusion (heat gain = heat loss)

1. What is temperature fusion?

Temperature fusion, also known as heat gain = heat loss, is a physical process in which two objects at different temperatures come into contact with each other and eventually reach a common temperature.

2. What is the significance of temperature fusion?

Temperature fusion is a fundamental concept in thermodynamics and plays a crucial role in understanding how heat is transferred between two objects. It also helps in predicting the final temperature of a system after the transfer of heat.

3. How does temperature fusion occur?

Temperature fusion occurs through the transfer of heat from one object to another. This transfer can happen through three modes of heat transfer: conduction, convection, and radiation.

4. What factors affect temperature fusion?

The rate of temperature fusion is affected by several factors, including the temperature difference between the two objects, the type of material of the objects, their surface area, and the duration of contact between them.

5. Can temperature fusion be reversed?

Yes, temperature fusion can be reversed through the process of cooling. When two objects with different temperatures come into contact, heat will flow from the warmer object to the cooler one until they reach a common temperature. If the objects are then separated, the heat flow will reverse, and the objects will return to their original temperatures.

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