A 0.3423 g sample of pentane, C5H12, was burned in a bomb calorimeter. The temperature of the calorimeter

and the 1.000 kg of water contained therein rose from 20.22°C to 22.82°C. The heat capacity of the
calorimeter is 2.21 kJ/°C. The heat capacity of water = 4.184 J/g·°C. What is the heat of combustion, in
kilojoules, per gram of pentane?

I'm confused on number 6 on the take home test. in this problem are we going to use the heat capacity of calorimeter is 2.21kJ/c. I'm confused on how to set this problem up
1)I converted 1.000kg in to grams= 1000g
2) I used q=m*c* the change in T
1000g*(4.184J/g* C)*(22.82-20.22) =10460 J
This is where I get stuck at and i'm not sure if i set the problem up right. can you explain to me if i'm doing it right, and/or show me how to set this problem up?

This sounds right to me! Next step to find kJ per gram, would be to convert your J to kJ, and divide by the amount of grams given. This is odd, however, since normally these problems want kJ per mole, not gram. I hope this helps!

It appears to me you have accounted for only a part of the heat.

q = [mass x specific heat x (Tfinal-Tintial)] + [Ccal*(Tfinal-Tinitial)]
The first term takes care of the heat gain in the water and the second term takes care of heating the calorimter.

To solve this problem, you need to calculate the heat released by the combustion of pentane and then divide it by the mass of pentane burned.

Here's how you can set up the problem step-by-step:

1) Calculate the heat released by the water:
q_water = mcΔT
where m is the mass of water and c is the specific heat capacity of water.

So, q_water = (1000g)(4.184 J/g°C)(22.82 - 20.22)°C
= 83480 J

2) Calculate the energy absorbed by the calorimeter:
q_calorimeter = C_calorimeterΔT_calorimeter
where C_calorimeter is the heat capacity of the calorimeter and ΔT_calorimeter is the change in temperature of the calorimeter.

So, q_calorimeter = (2.21 kJ/°C)(22.82 - 20.22)°C
= 5.7422 kJ

3) Calculate the total heat absorbed in the system:
q_total = q_water + q_calorimeter
q_total = 83480 J + 5.7422 kJ
= 59480 J + 5742.2 J
= 65222.2 J

4) Convert the heat absorbed to kilojoules:
q_total = 65222.2 J / 1000
= 65.222 kJ

5) Calculate the mass of pentane burned:
Given mass of pentane = 0.3423 g

6) Calculate the heat of combustion per gram of pentane:
heat of combustion = q_total / mass of pentane

heat of combustion = 65.222 kJ / 0.3423 g
= 190.49 kJ/g

Therefore, the heat of combustion per gram of pentane is 190.49 kJ/g.

To solve this problem, you are on the right track. Let's go step by step to set up and solve the problem accurately.

Step 1: Convert 1.000 kg to grams:
1.000 kg = 1000 g (which you already did correctly).

Step 2: Calculate the heat transferred to the water:
Use the formula q = m * c * ΔT, where
q is the heat transferred,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.

Given:
m = 1000 g (mass of the water)
c = 4.184 J/g·°C (specific heat capacity of water)
ΔT = 22.82°C - 20.22°C = 2.6°C

Plug in the values into the formula:
q = 1000 g * 4.184 J/g·°C * 2.6°C
q = 10,854.4 J (approximately)

So, the heat transferred to the water is 10,854.4 J.

Step 3: Calculate the heat transferred to the calorimeter:
Use the formula q = m * c * ΔT, where
q is the heat transferred,
m is the mass of the calorimeter,
c is the specific heat capacity of the calorimeter, and
ΔT is the change in temperature.

Given:
m = unknown (mass of the calorimeter)
c = 2.21 kJ/°C (specific heat capacity of the calorimeter)
ΔT = 2.6°C (we assume both the water and calorimeter have the same ΔT)

We don't know the mass of the calorimeter, so let's assume it as "x" grams.

q = x g * 2.21 kJ/°C * 2.6°C
q = 5.7462 x kJ (approximately), where "x" represents the unknown mass of the calorimeter

Step 4: Total heat transferred to the system:
The total heat transferred to the system (water + calorimeter) is the sum of the heat transferred to the water and the heat transferred to the calorimeter.

Total q = q_water + q_calorimeter
Total q = 10,854.4 J + 5746.2 x kJ (since q_calorimeter is in kJ)

Step 5: Convert all units to match for comparison:
We noticed that q_water is given in J, whereas q_calorimeter is in kJ. To compare the units, let's convert q_water to kJ.

10,854.4 J ÷ 1000 = 10.8544 kJ

So, Total q = 10.8544 kJ + 5746.2 x kJ

Step 6: Set up the equation and solve for "x":
The total heat transferred to the system is equal to the heat released during the combustion of pentane.

Total q = Heat of combustion * mass of the pentane
10.8544 kJ + 5746.2 x kJ = Heat of combustion * 0.3423 g

Now, rearrange the equation to solve for "x":
Heat of combustion = (10.8544 kJ + 5746.2 x kJ) / 0.3423 g

Step 7: Calculate the heat of combustion per gram of pentane:
Plug in the values and solve for the heat of combustion per gram of pentane.

Heat of combustion = (10.8544 kJ + 5746.2 x kJ) / 0.3423 g

This will give you the heat of combustion per gram of pentane in kilojoules.