Calorimetry - Calculating absorbed and released heat

In summary, the experiment involved burning ethanol in an alcohol burner under a can of water placed in a larger can. The heat absorbed by the water was calculated to be 35.1kJ and the molar heat of combustion of ethanol was determined to be 81.08kJ/mol. The difference between Q and "delta"H depends on the specific circumstances and definition of enthalpy."
  • #1
edimeo25
6
0

Homework Statement


Calculate the heat absorbed by the water in the calorimeter and the heat released by 1,0g of burning ethanol. Then, calculate the molar heat of combustion (kJ/mol) of the ethanol.

The calorimeter used was very makeshift: ethanol in an alcohol burner was lit under a can containing water, both of which were placed inside a hollowed out larger can.

Here is what is known:
mass of ethanol burned = 20g
mass of heated water = 280g
change in temperature = 30"degrees"C


Homework Equations



Q(water) = -Q(ethanol)
mc"delta"T = - (mc"delta"T)
280g*4.184J/g"degree"C*30 = -Q(ethanol)

The Attempt at a Solution


Q(water) = mc"delta"T = 280g*4.184J/g"degree"C*30"degrees"C = 35145.6J = 35.1kJ

Q(ethanol) = -Q(water) = -35.1kJ (for 20g)

Q(per gram of ethanol) = -35.1kJ / 20g = -1.755kJ/g = -1.76kJ/g

Q or "delta"H? (per mol of ethanol) = -1.76kJ/g * 46.07g/mol = 81.08kJ/mol


--> Does this make sense? Is it right? And what is the difference between Q and "delta"H?
 
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  • #2
Looks OK.

Check enthalpy definition. In some circumstances it is identical to Q, in some it is not.

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  • #3


Firstly, your calculations and approach seem correct. The difference between Q and "delta"H is that Q represents the total amount of heat absorbed or released in a process, while "delta"H represents the change in enthalpy. Enthalpy is a thermodynamic quantity that takes into account not only the heat absorbed or released, but also changes in pressure and volume. In this case, the heat absorbed by the water and the heat released by the burning ethanol are equal in magnitude but opposite in sign, resulting in a net change of zero in enthalpy. The molar heat of combustion, which you calculated to be 81.08 kJ/mol, is a measure of how much heat is released when one mole of ethanol is burned. Your result seems reasonable and in line with known values for the molar heat of combustion of ethanol. However, it is always a good idea to double check your calculations and units to ensure accuracy. Overall, your response shows a good understanding of the principles of calorimetry and thermodynamics.
 

Related to Calorimetry - Calculating absorbed and released heat

1. What is calorimetry?

Calorimetry is the scientific study of measuring heat and its transfer from one substance to another. It involves using a calorimeter, a device that measures the heat absorbed or released during a chemical reaction or physical process.

2. How is heat released or absorbed calculated in calorimetry?

In calorimetry, the heat released or absorbed is calculated using the equation Q = mcΔT, where Q is the amount of heat, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

3. What are some applications of calorimetry in scientific research?

Calorimetry is used in a variety of scientific fields, such as chemistry, physics, and biology. It can be used to study chemical reactions, determine the energy content of foods, and measure the specific heat capacity of materials.

4. How does a calorimeter work?

A calorimeter typically consists of a container filled with a known mass of water, an insulated chamber to contain the reaction or process being studied, and a thermometer to measure the temperature change. The heat released or absorbed by the reaction or process is then calculated using the change in temperature of the water.

5. What are the limitations of using calorimetry to measure heat?

While calorimetry can provide accurate measurements of heat, there are some limitations to consider. For example, it assumes that all the heat produced by a reaction or process is absorbed by the water, which may not always be the case. It also does not take into account any heat lost to the surroundings, which can affect the accuracy of the results.

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