Steps for this problem got me confused

Question

Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 17.0 °C are mixed. The specific heat of water is 4.184 ^J/_g°C.

Answer 1

For this question, we will use a slightly modified version of the equation Q = m*c*ΔT. The mass (m) and the c (ae calculated just like they are in any other Q = m*c*ΔT problem. The difference occurs with how we calculate ΔT. Let's first draw a number line and put on it the two temperatures we are given in the problem (100 degrees Celsius and 17.0 degrees Celsius):

<------- 17.0 ---------------------------------- 100 ----->

We are asked to find the final temperature. Let's call the final temperature "x". We don't know the value of x, but we can make educated guesses about it. It will be above 17.0 degrees Celsius and below 100 degrees Celsius. So we put x somewhere between 17.0 and 100 on the number line like this:

<------- 17.0 ------------- x -------------------- 100 ----->

Now let's find the formula for ΔT for the 50.00 gram sample of water. Remember that we write ΔT differently for questions that involve two mixtures. To write ΔT for the 50.00 gram sample of water, we look at the initial temperature for this sample (100 degrees Celsius) and x (the final temperature). For this kind of question,

ΔT = larger number -- smaller number. Let's look back at our number line. When we do, we see that 100 is bigger than x. So for the 50.00 gram sample of water, ΔT = 100 -- x. Now we can write the other information we know for the 50.00 gram sample of water:

m = 50.00 g (we don't have to change this to kg since the specific heat capacity for water that was given to use has the gram in its units)

c = 4.184 J/ (g*C)

ΔT = 100 -- x

So for this 50.00 g sample of water, Q = 50 * 4.184 * (100--x)

Now let's figure out ΔT for the 100 gram sample of water. To write ΔT for the 100 gram sample of water, we look at the initial temperature for this sample (17.0 degrees Celsius) and x (the final temperature). For this kind of question,

ΔT = larger number -- smaller number. Let's look back at our number line. When we do, we see that x is bigger than 17.0. So for the 100 gram sample of water, ΔT = x -- 17.0. Now we can write the other information we know for the 100 gram sample of water:

m = 100 g

c = 4.184 J/(g*C)

ΔT = x -- 17.0

So for the 100 gram sample of water, Q = 100 * 4.184 * (x -- 17.0)

Remember that the equation for the 50.00 gram sample of water is Q = 50 * 4.184 * (100--x) . Now we put the equation for the 50.00 gram sample of water equal to the equation for the 100 g sample of water. We omit the Qs. So it looks like this:

50 * 4.184 * (100 -- x) = 100 * 4.184 * (x -- 17.0)

Now we start solving for x (the final temperature) using algebra. Note that both sides have a 4.184 in the equation. We can divide both sides by 4.184, and get rid of the 4.184. So the equation looks like this:

50 * (100 -- x) = 100 * (x -- 17.0)

Now notice that there is a 50 on the left side of the equals sign and a 100 on the right side. If we divide both sides by 50, we get rid of the 50 on the left side. On the right side, 100 / 50 = 2. So our equation looks like this:

(100 -- x) = 2 * (x -- 17.0)

We distribute the 2 across the (x -- 17.0) so the equation looks like this:

100 -- x = 2x -- 34

Now we simplify more until we get:

134 = 3x

So x = 134 / 3 = 44.667. Since x is the final temperature, the unit is Celsius.

So the final temperature is 44.7 degrees Celsius