When methane are burned in a bomb calorimeter the temperature rises from 21.00℃ to 25.18℃. The heat capacity of methane is 3.754x10^3 J/℃. The amount of heat, q (in kJ) absorbed by the calorimeter apparatus is:
Correct Answer: 18.68 kJ
My Work and my answer:
q=(16.042g)(3.754x10^3 J/℃)(25.18℃-21.00℃)
q=251726.57J
Frankly, I don't think you have copied the problem correctly. Where did the 16.042 in your work come from?
To calculate the amount of heat absorbed by the calorimeter apparatus, you need to use the formula:
q = m * c * ΔT
Where:
q = amount of heat absorbed (in joules or J)
m = mass of the substance (in grams or g)
c = heat capacity of the substance (in J/℃)
ΔT = change in temperature (in ℃)
In this case, the substance is methane and its mass is given as 16.042g. The heat capacity of methane is 3.754x10^3 J/℃. The change in temperature is from 21.00℃ to 25.18℃.
Plugging in the values into the formula, we get:
q = (16.042g) * (3.754x10^3 J/℃) * (25.18℃ - 21.00℃)
Calculating this out:
q = (16.042g) * (3.754x10^3 J/℃) * (4.18℃)
q = 251732.232 J
Now, to convert this to kilojoules (kJ), we divide by 1000:
q = 251.732232 kJ
So, the correct answer for the amount of heat absorbed by the calorimeter apparatus is approximately 251.73 kJ, rounded to two decimal places.