Equilibrium Temp of two fluids

In summary, the conversation discusses the problem of finding the final equilibrium temperature in degrees Celsius when a 5-gallon container of water at 212 degrees Fahrenheit is added to 50 gallons of water at 50 degrees Fahrenheit. The relevant equation used is Qlost = Qgained. The incorrect use of the equation m1C∆T1 = m2C∆T2 initially led to an incorrect final temperature of 0 degrees Celsius, but was corrected to 32 degrees Fahrenheit or 0 degrees Celsius. The mistake was in not accounting for the negative and positive signs of the heat transfer.
  • #1
trying_mybest
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4
Homework Statement
A 5-gallon container of water (1 gal = 3.79 liter) at 212 degrees F is added to 50 gallons of water at 50 degrees F. What is the final equilibrium temperature in degrees C?
Relevant Equations
Qlost = Qgained
Not really sure how to start this one.
 
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  • #2
trying_mybest said:
Homework Statement:: A 5-gallon container of water (1 gal = 3.79 liter) at 212 degrees F is added to 50 gallons of water at 50 degrees F. What is the final equilibrium temperature in degrees C?
Relevant Equations:: Qlost = Qgained

Not really sure how to start this one.
If the final temperature is T degrees F, what are the values of Qlost and Qgained?
 
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  • #3
Final temp should be ˚C.

I used m1C∆T1 = m2C∆T2

I keep getting 0 for the final temp when it's supposed to be 18 C
 
  • #4
trying_mybest said:
I keep getting 0 for the final temp when it's supposed to be 18 C
And you'd like someone to guess where you are going wrong?
trying_mybest said:
Final temp should be ˚C.
Yes, but it will be simpler to do one conversion at the end than two at the start.
 
  • #5
trying_mybest said:
Final temp should be ˚C.

I used m1C∆T1 = m2C∆T2

I keep getting 0 for the final temp when it's supposed to be 18 C
That equation looks fine but what do you put for ##\Delta T_1## and ##\Delta T_2##?
 
  • #6
m1 = 5 gal * 3.79 L/gal * 1000 g/L = 18,750 g
m2 = 50 gal * 3.79 L/gal * 1000 g/L = 187,500 g

m1*C*(Tf - T1) = m2*C*(Tf - T2)

m1CTf - m1CT1 = m2CTf - m2CT2

m1CTf - m2CTf = m1CT1 - m2CT2
Tf(m1 - m2) = m1T1 - m2T2
Tf = (m1T1 - m2T2) / (m1 - m2)
Tf = (18,750*212 - 187,500*50) / (18,750 - 187,500)
Tf = -5,400,000 / -168,750 = 32 ˚F = 0 ˚C
 
  • #7
Well, i think the mistake lies right at the start.
$$m_1C(T_f-T_1)=m_2C(T_f-T_2)$$ is not correct, the correct is $$-m_1C(T_f-T_1)=m_2C(T_f-T_2)$$ because one body is losing heat (so the heat will be negative) and the other is gaining heat (so the heat will be positive).
 
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  • #8
Delta2 said:
Well, i think the mistake lies right at the start.
$$m_1C(T_f-T_1)=m_2C(T_f-T_2)$$ is not correct, the correct is $$-m_1C(T_f-T_1)=m_2C(T_f-T_2)$$ because one body is losing heat (so the heat will be negative) and the other is gaining heat (so the heat will be positive).

Thank you! I caught that mistake earlier, fixed it, and got the correct answer.
 
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1. What is the equilibrium temperature of two fluids?

The equilibrium temperature of two fluids is the temperature at which the two fluids, when combined, reach a stable and balanced temperature. This means that the rate of heat transfer between the two fluids is equal, resulting in no further change in temperature.

2. How is the equilibrium temperature of two fluids calculated?

The equilibrium temperature of two fluids can be calculated using the formula: Teq = (m1 x c1 x T1 + m2 x c2 x T2) / (m1 x c1 + m2 x c2), where Teq is the equilibrium temperature, m is the mass of the fluid, c is the specific heat capacity, and T is the initial temperature of each fluid.

3. What factors affect the equilibrium temperature of two fluids?

The factors that affect the equilibrium temperature of two fluids include the initial temperatures of the two fluids, their masses, and their specific heat capacities. Other factors such as the rate of heat transfer and the presence of external heat sources can also impact the equilibrium temperature.

4. How does the equilibrium temperature of two fluids relate to heat transfer?

The equilibrium temperature of two fluids is directly related to heat transfer. When the two fluids are at different initial temperatures, heat will transfer from the hotter fluid to the cooler fluid until they reach equilibrium. The rate of heat transfer will decrease as the two fluids approach equilibrium.

5. Can the equilibrium temperature of two fluids change?

Yes, the equilibrium temperature of two fluids can change if there is a change in any of the factors that affect it, such as the initial temperatures, masses, or specific heat capacities of the fluids. It can also change if there is a change in the rate of heat transfer or the presence of external heat sources.

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