Solving Algebraically for T2: Q Lost = Q Gained

  • Thread starter Thread starter Soliloquy12
  • Start date Start date
  • Tags Tags
    Lost
AI Thread Summary
The discussion revolves around solving for T2 in a heat transfer problem using the equation QLost = QGained. The user derived an equation for T2 but encountered discrepancies in their results, obtaining T2 = 123.66°C from the original equation and T2 = 76.34°C from their derived equation. Other participants pointed out that the final temperature should logically fall between the two initial temperatures, suggesting a possible algebraic error in the user's calculations. They recommended clarifying the equation by adding parentheses for accuracy and urged the user to share detailed steps to identify the mistake. The conversation highlights the importance of careful algebraic manipulation in solving thermal equilibrium problems.
Soliloquy12
Messages
2
Reaction score
0

Homework Statement



Solve algebraically for T2 then find T2.

T1C=63 Celsius
T1H=79.5 Celsius
MCW=8.63g
MBW=36.48g

Homework Equations



QLost = QGained

-mbwcw(T2-T1H) = mcwcw(T2-T1C)


The Attempt at a Solution



Trying to solve algebraically I arrived at:

T2 = mcwt1C+mbwt1h/(mbw+mcw)

But when substituting the values I get T2 = 123.66J from the original equation and T2 = 76.34 when using the equation I derived from the original.

Can someone please show me where I am going wrong with the equation here?

Thanks!
 
Physics news on Phys.org
Soliloquy12 said:
But when substituting the values I get T2 = 123.66J from the original equation
What do you mean? How exactly did you get that result? (I assume that's the final temperature of some mixture, not an energy in Joules.)
and T2 = 76.34 when using the equation I derived from the original.
I didn't do the calculation, but if the problem is what I think it is, that sounds reasonable.
 
Yes sorry i meant degrees celsius. I substitutted the values into the initial "long" equation when I got 123.66 degrees celsius.
 
Soliloquy12 said:
Yes sorry i meant degrees celsius. I substitutted the values into the initial "long" equation when I got 123.66 degrees celsius.
I still don't know what you mean. The initial "long" equation is the same equation that you rearranged to solve for T2. All you did was solve it algebraically--it's still the same equation.

Show exactly what you did. What values did you substitute?
 
Soliloquy12 said:

Homework Statement





-mbwcw(T2-T1H) = mcwcw(T2-T1C)


The Attempt at a Solution



Trying to solve algebraically I arrived at:

T2 = mcwt1C+mbwt1h/(mbw+mcw)

But when substituting the values I get T2 = 123.66J from the original equation and T2 = 76.34 when using the equation I derived from the original.

Can someone please show me where I am going wrong with the equation here?

Thanks!
First, I would suggest that you put parentheses around the two terms in your numerator for clarity and to prevent mistakes.

Second, I'm assuming that this problem involves adding something hot to something cool because you didn't really say. In problems such as that the final temperature will lie between the two starting temperatures. Therefore you should suspect that you made a simple algebraic error when you computed 123.66 degrees. As Dr AL indicated, without seeing the details of your calculation, one can't say more.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top