# Solving Algebraically for T2: Q Lost = Q Gained

• Soliloquy12
In summary, the problem involves solving for T2 in an equation involving temperature changes and the specific heat capacity of two substances. When substituting the given values into the original equation, T2 is calculated to be 123.66 degrees Celsius, but when using the equation derived from the original, T2 is calculated to be 76.34 degrees Celsius. The final temperature should lie between the two starting temperatures, so there is likely a simple algebraic error in the calculation.
Soliloquy12

## 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!

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.

## 1. How do I solve for T2 in "Q Lost = Q Gained" using algebra?

To solve for T2 in this equation, you will need to isolate the variable by performing inverse operations on both sides. Begin by subtracting Q Gained from both sides to get "Q Lost - Q Gained = 0". Then, divide both sides by the coefficient of T2 to get "T2 = (Q Lost - Q Gained)/coefficient". This will give you the value of T2 that satisfies the equation.

## 2. Can I solve for T2 using any algebraic method?

Yes, there are multiple algebraic methods that can be used to solve for T2 in this equation. The most common method is to use the properties of equality to isolate the variable, as described in the previous answer. Other methods include substitution and elimination, which may be more useful in certain situations.

## 3. What does the equation "Q Lost = Q Gained" represent?

This equation represents the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. In this case, "Q Lost" represents the amount of energy lost from a system, while "Q Gained" represents the amount of energy gained by the system.

## 4. How does solving for T2 help in understanding energy transfer?

By solving for T2 in this equation, you can determine the specific amount of energy that is lost or gained in a given system. This can help in understanding the efficiency of energy transfer and identifying any potential issues or improvements in the system.

## 5. Can this equation be applied to other scientific principles besides energy transfer?

Yes, the principle of conservation of energy can be applied to various other scientific principles, such as mass and momentum. In these cases, the equation may be modified to incorporate other variables, but the overall concept of energy conservation remains the same.

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