If the heat from burning 6.000 g of

C6H6 2C6H6(l) + 15O2(g) -> 12CO2(g) + 6H2O(l) + 6542 kJ is added to 5691 g of water at 21 °C, what is the final temperature of the water?
Please tell me what I am doing wrong:
293,551= 5691*4.2*(Tf-21C)
293551= 23902.2Tf - 50946.2
343946.2= 23902.2Tf
14.4=Tf

The homework program is telling me it's wrong.

There seems to be an error in your calculation. Let's go through the steps again:

First, we need to calculate the heat gained by the water using the formula: q = m * c * ΔT, where:
- q is the heat gained by the water,
- m is the mass of the water (5691 g),
- c is the specific heat capacity of water (4.18 J/g°C), and
- ΔT is the change in temperature (Tf - 21°C).

q = m * c * ΔT

Plugging in the values:
q = 5691 g * 4.18 J/g°C * (Tf - 21°C)

We need to convert the heat from kJ to J:
6542 kJ = 6542000 J

Now, equating the heat gained by the water to the heat released by the burning of C6H6:
6542000 J = -6542000 J (since heat is released by the reaction)

Now, let's solve for Tf:
5691 g * 4.18 J/g°C * (Tf - 21°C) = -6542000 J

Multiply the numbers out:
23751.38 g°C * (Tf - 21°C) = -6542000 J

Now, rearrange the equation to solve for Tf:
Tf - 21°C = -6542000 J / 23751.38 g°C
Tf - 21°C = -275.41 g°C/J

Add 21°C to both sides:
Tf = -275.41 g°C/J + 21°C
Tf = -275.41 g°C/J + 489.98°C

After evaluating the expression, the final temperature of the water comes out to be approximately 214.57°C.

Therefore, the error in your calculation was in your substitution of values and division. Please double-check your calculations.

To solve this problem, you correctly identified that you need to use the heat capacity formula:

q = m * c * ΔT,

where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

You correctly substituted the values into the equation:

293,551 = 5691 * 4.2 * (Tf - 21).

However, it seems like there might be a calculation error in your calculations. Let's correct it:

First, multiply the mass of water (5691 g) by the specific heat capacity of water (4.2 J/g°C):

Q1 = 5691 g * 4.2 J/g°C = 23,870.2 J.

Since the heat released by the burning of C6H6 is 6542 kJ, we need to convert it to joules:

Q2 = 6542 kJ * 1000 J/kJ = 6,542,000 J.

Next, add Q1 and Q2 together to find the total heat absorbed by the water:

Q = Q1 + Q2 = 23,870.2 J + 6,542,000 J = 6,565,870.2 J.

Now, rearrange the heat capacity formula to solve for ΔT:

Q = m * c * ΔT.

ΔT = Q / (m * c).

Plug in the values:

ΔT = 6,565,870.2 J / (5691 g * 4.2 J/g°C) ≈ 244.2 °C.

Finally, to find the final temperature, add the change in temperature to the initial temperature:

Tf = 21 °C + 244.2 °C ≈ 265.2 °C.

So, the final temperature of the water should be approximately 265.2 °C.

Please double-check your calculations to ensure accuracy.