In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23 degrees Celsius. If 4.00 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution of CaCl2 is -82.8 kJ/mol.

Here are my numbers:

100mL water, Temp 1: 23C, Temp2: ?, 7.20 g CaCl2, DeltaH=-82.8kJ/mol.

First:
Temperature is going to increase, so turn your -82.8kJ/mol to +82.8.
Also, 100 mL water = 100 g water b/c water's density is 1g/mL

Second:
82.8kJ/mol=82,800J/mol

Third:
Molar mass of CaCl2=110.9 g/mol

Fourth:
(82,800 J/mol)(1mol/110.98g)=746.08 J/g

Fifth:
746.08J/g * 7.20 g = 5371.78 J

Sixth:
(100g + 7.2 g total mass)(4.18 specific heat)(T2-23C)=5371.78J

T2-23=12
T2=35C

HOpe this helps!

Wow finally someone who actually was able to help me get the right answer!!! ^ (I'm not sure where you got 7.2g for CaCl2, but apart from that you did everything right!)

Thanks for the help!

Well, it seems like we have a hot topic here! Let's dive into this coffee-cup calorimeter experiment with a splash of humor, shall we?

First, let's calculate the number of moles of CaCl2. We have 4.00 g of it. So, assuming CaCl2 is really clingy, we need to handle it with care and calculate its molar mass. It turns out that the molar mass of CaCl2 is approximately 111 g/mol. Therefore, we have about 0.036 moles of CaCl2 (4.00 g / 111 g/mol).

Now, let's talk about the heat of solution. CaCl2 is a bit of a drama queen, and it releases energy when it dissolves. The heat of solution is -82.8 kJ/mol. We multiply this by the number of moles of CaCl2 we have to get the energy released: -82.8 kJ/mol * 0.036 mol = -2.98 kJ.

Now, this is where things start to heat up (pun intended). We can use the heat gained by the water to calculate the final temperature. The heat gained or lost is given by the equation Q = mcΔT, where Q is the energy gained or lost, m is the mass of the water, c is the specific heat capacity of water (which is about 4.18 J/g·°C), and ΔT is the change in temperature.

We know the mass of the water is 100 g (since 100 mL of water is approximately 100 g). We already have the energy released, which is -2.98 kJ, but we need to convert it to joules because the specific heat capacity is in joules. So -2.98 kJ * 1000 J/1 kJ = -2980 J.

We rearrange the equation Q = mcΔT and solve for ΔT: ΔT = Q / (m * c). Plugging in the values, we have ΔT = -2980 J / (100 g * 4.18 J/g·°C). Calculating this gives us ΔT ≈ -7.13 °C.

But wait, we're not heating things, we're cooling them down! So we need to make ΔT positive, giving us a final temperature of 23 °C - 7.13 °C ≈ 15.87 °C.

So, in this experiment, the solution in the calorimeter is expected to reach a final temperature of approximately 15.87 °C. Now, that's a cool outcome!

To find the final temperature of the solution in the calorimeter, we need to use the concepts of heat transfer and heat capacity.

First, let's calculate the heat released or absorbed when the CaCl2 is dissolved in water. The heat of solution is given as -82.8 kJ/mol. Since we have 4.00 g of CaCl2, we need to convert it to moles.

To calculate the moles of CaCl2, we divide the given mass by its molar mass. The molar mass of CaCl2 is the sum of the molar masses of calcium (Ca) and chlorine (Cl). From the periodic table, we find that the molar mass of Ca is 40.08 g/mol and the molar mass of Cl is 35.45 g/mol.

So, molar mass of CaCl2 = 40.08 g/mol + (2 * 35.45 g/mol) = 110.98 g/mol

Now, we can calculate the number of moles of CaCl2:
moles of CaCl2 = (4.00 g) / (110.98 g/mol)

Next, we can calculate the heat released or absorbed:

Q = moles of CaCl2 * heat of solution
Q = (moles of CaCl2) * (-82.8 kJ/mol)

Now, let's determine the heat capacity of the calorimeter (Ccal). The heat capacity is the amount of heat required to raise the temperature of a substance by 1 degree Celsius. It depends on the mass and specific heat capacity of the substance.

In this case, the substance is water, which has a specific heat capacity of 4.18 J/g*C and a mass of 100 mL (or 100 g). Thus, the heat capacity of the water in the calorimeter is:

Ccal = (mass of water in the calorimeter) * (specific heat capacity of water)
Ccal = (100 g) * (4.18 J/g*C)

Now, we can use the formula for heat transfer:

Q = Ccal * (change in temperature)

Since the initial temperature of the calorimeter is 23 degrees Celsius, and we want to find the final temperature, we can rewrite the formula as:

Q = Ccal * (final temperature - initial temperature)

Now, let's plug in the known values:

Q = Ccal * (final temperature - 23 degrees Celsius)

We already calculated Q and Ccal. Now we need to rearrange the equation to solve for the final temperature:

(final temperature - 23 degrees Celsius) = Q / Ccal

Now, let's substitute the values for Q and Ccal:

(final temperature - 23 degrees Celsius) = (moles of CaCl2 * (-82.8 kJ/mol)) / ((100 g) * (4.18 J/g*C))

(final temperature - 23 degrees Celsius) = (moles of CaCl2 * (-82.8 kJ/mol)) / (418 J/C)

Now, we have everything we need to find the final temperature. Plug in the value for moles of CaCl2 that we calculated earlier, and solve for the final temperature.

I hope this explanation helps you understand how to calculate the final temperature of the solution in the calorimeter!

^^ yes this is how you get to the right answer, if you were given a dif temp just plug it in and solve.