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)*(295.97 K -293.37 K) =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?

You need to pay attention to your units. The temperature for heat capacity is in given with C not K. You have to use C as temperature and not K. Also, I have no idea what the problem is asking; all I see is numbers, so I really can't check, nor can anyone else.

my professor told us to change them the temperture from C to K but I will be sure to remember that. but the problem is: 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?

To solve this problem, it's important to understand the concept and formula for calculating heat (q) in a calorimetry experiment. The formula is:

q = m * c * ΔT

Where:
q = heat transferred (in J)
m = mass of the substance being heated or cooled (in grams, in this case)
c = specific heat capacity of the substance (in J/g·°C)
ΔT = change in temperature (in °C)

In this problem, it seems that you are trying to calculate the heat transferred when heating a 1.000 kg (or 1000 g) sample using a calorimeter with a heat capacity of 2.21 kJ/°C.

To set up the problem correctly, you need to account for the heat absorbed by both the sample and the calorimeter. The heat absorbed by the calorimeter can be calculated using the formula:

q_calorimeter = C_calorimeter * ΔT

Where:
q_calorimeter = heat absorbed by the calorimeter
C_calorimeter = heat capacity of the calorimeter (in J/°C)
ΔT = change in temperature of the calorimeter

Since the heat capacity of the calorimeter is given as 2.21 kJ/°C, you need to convert it to J/°C by multiplying by 1000:

C_calorimeter = 2.21 kJ/°C * 1000 J/1 kJ = 2210 J/°C

Now you can calculate q_calorimeter using the given values:

q_calorimeter = 2210 J/°C * (295.97°C - 293.37°C) = 5782 J

Next, you need to calculate the heat absorbed by the sample using the first formula:

q_sample = m * c * ΔT

Plugging in the values:

q_sample = 1000 g * 4.184 J/g·°C * (295.97°C - 293.37°C) = 10,473 J

Finally, to get the total heat transferred, you need to sum up the heat absorbed by the sample and the calorimeter:

q_total = q_sample + q_calorimeter = 10,473 J + 5782 J = 16,255 J

Therefore, the correct answer is 16,255 J for the total heat transferred in this experiment.

Remember to always double-check your conversions and units when solving these types of problems.