The molar enthalpy of combustion of glucose is -2803 kJ. A mass of 1.000 g glucose is combusted in a bomb calorimeter. If the calorimeter contains 875 g H2O and the bomb has a heat capacity of 457 J/C, what is the temperature increase of the bomb calorimeter? the specific heat capacity of water is 4.184 J/g*K and the molar mass of glucose is 180.2 g/mol.
See your other post where I responded.
To find the temperature increase of the bomb calorimeter, we need to calculate the heat released during the combustion of glucose and then use that to find the temperature increase.
First, let's calculate the heat released during the combustion of 1.000 g of glucose. We can use the molar enthalpy of combustion to do this.
The molar enthalpy of combustion of glucose is given as -2803 kJ/mol. We can convert this to J/mol by multiplying by 1000 since 1 kJ is equal to 1000 J.
-2803 kJ/mol * 1000 J/kJ = -2,803,000 J/mol
Now, let's convert this to J/g since we only have 1.000 g of glucose.
-2,803,000 J/mol / 180.2 g/mol = -15,546.95 J/g
This means that when 1.000 g of glucose is combusted, 15,546.95 J of heat is released.
Next, we need to calculate the heat capacity of the bomb calorimeter.
The heat capacity of the bomb calorimeter is given as 457 J/C. This means that for every 1 degree Celsius increase in temperature, the bomb calorimeter absorbs 457 J of heat.
Now, we need to calculate the amount of heat absorbed by the bomb calorimeter when the temperature increases.
We know that the bomb calorimeter contains 875 g of water, and the specific heat capacity of water is given as 4.184 J/g*K.
The formula to calculate the heat absorbed is:
Heat absorbed = mass of water * specific heat capacity of water * temperature increase
Substituting the values into the formula:
Heat absorbed = 875 g * 4.184 J/g*K * ΔT
Since the heat released during combustion is equal to the heat absorbed by the bomb calorimeter, we can set the two equations equal to each other:
-15,546.95 J/g = 875 g * 4.184 J/g*K * ΔT
Simplifying the equation:
-15,546.95 J/g = 3655.3 g*J/K * ΔT
Next, solve for ΔT:
ΔT = -15,546.95 J / (3655.3 g*J/K) ≈ -4.25 K
The negative sign indicates that the temperature decreases instead of increasing. However, it's important to note that the bomb calorimeter can absorb heat from its surrounding and the surrounding can provide that. So, the final answer would depend on the initial temperature of the bomb calorimeter and its surroundings.