Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
Answer:
The person who just wanted points is annoying
Explanation:
WHY WOULD YOU DO THAT
C6H12O + 6OC2 + 6H2O + energy
Answer: mmmmmm asking for mrs.howard work I see lol good luck grace
Explanation:
Heat energy is the amount of heat there is in a substance. This may be cold heat or warm heat. Temperature is how hot or cold a substance is. So because there is a higher volume of heat in the iceberg, we say it has more heat energy.
