The correct equation for calculating the equilibrium temperature of two fluids is derived from the principle of conservation of energy, specifically using the formula -m1C(Tf - T1) = m2C(Tf - T2). In this discussion, a 5-gallon container of water at 212°F is mixed with 50 gallons at 50°F, leading to a final equilibrium temperature of 18°C. The mistake in the initial approach was not accounting for the sign of heat transfer, where one body loses heat and the other gains it. The final calculation confirms that the equilibrium temperature is indeed 18°C.
PREREQUISITESStudents in physics or engineering courses, educators teaching thermodynamics, and anyone involved in practical applications of heat transfer principles.
If the final temperature is T degrees F, what are the values of Qlost and Qgained?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.
And you'd like someone to guess where you are going wrong?trying_mybest said:I keep getting 0 for the final temp when it's supposed to be 18 C
Yes, but it will be simpler to do one conversion at the end than two at the start.trying_mybest said:Final temp should be ˚C.
That equation looks fine but what do you put for ##\Delta T_1## and ##\Delta T_2##?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
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).