The heat of combustion of benzene, C6H6, is -41.74 kJ/g. Combustion of 3.65 of benzene causes a temperature rise of 4.41 C^o in a certain bomb calorimeter. What is the heat capacity of this bomb calorimeter? (kJ/C^o)

So, I what I did was I multiplied -41.74 kJ/g and 3.65 to get the kJ alone. This gave me -152.351. Then, I divided that number by -4.41 (negative because it goes down by that amount) to get kJ/C^o. I got 34.5 kJ/C^o as my answer. Apparently, this is not the answer. Help?

yesh

If you carry that out to one more place MAYBE it is

(41.74*3.65/4.41) = 34.5467 and if I round that to 34.55 and round that to 34.6 to three significant figures; otherwise, your procedure looks ok to me.

To calculate the heat capacity of the bomb calorimeter, you need to use the equation:

q = m * C * ΔT

where:
q is the heat transferred (in this case, the heat of combustion of benzene),
m is the mass of the substance being combusted (in this case, benzene) in grams,
C is the heat capacity of the bomb calorimeter (what we need to find), and
ΔT is the change in temperature of the system (in this case, the temperature rise of the bomb calorimeter).

Using the given information:
Heat of combustion of benzene (q) = -41.74 kJ/g
Mass of benzene (m) = 3.65 g
Temperature rise (ΔT) = 4.41°C

Substituting the values into the equation, we have:

-41.74 kJ/g * 3.65 g = C * 4.41°C

Simplifying the equation:

-152.051 kJ = C * 4.41°C

Now, to solve for C (heat capacity), divide both sides of the equation by 4.41°C:

C = -152.051 kJ / 4.41°C

C ≈ -34.51 kJ/°C

So, the heat capacity of the bomb calorimeter is approximately -34.51 kJ/°C, not 34.5 kJ/°C.

To solve this problem, we need to apply the concept of heat capacity and use the known values provided in the question.

First, let's calculate the total heat released by the combustion of benzene. We know that the heat of combustion of benzene is -41.74 kJ/g, and we have 3.65 g of benzene. Therefore, the total heat released is:

Total Heat Released = Heat of Combustion × Mass of Benzene
= -41.74 kJ/g × 3.65 g
= -152.031 kJ (rounded to three decimal places)

Next, we need to calculate the heat capacity of the bomb calorimeter. The temperature rise caused by the combustion of 3.65 g of benzene is 4.41 °C. We can use the formula:

Heat Capacity = Total Heat Released / Temperature Rise

Heat Capacity = -152.031 kJ / 4.41 °C
= -34.463 kJ/°C (rounded to three decimal places)

The negative sign indicates a convention used to show that heat is being released by the reaction. However, it is not typically used when expressing heat capacity.

Therefore, the correct heat capacity of the bomb calorimeter in this case is approximately 34.463 kJ/°C.