Work Done ON or BY the System [Conceptual Question]

[CallMeShady]

Homework Statement

A student performs a combustion experiment by burning a mixture of fuel and oxygen in a constant-volume metal can surrounded by a water bath. During the experiment, the temperature of the water is observed to rise. Regard the mixture of fuel and oxygen as the system.
a) Has heat been transferred to/from the system? How can you tell?
b) Has work been done by/on the system? How can you tell?
c) What is the sign of ΔU of the system? How can you tell?
d) 1g of fuel is placed in the can for combustion. During combustion, the 200mL water bath is observed to increase in temperature by 35.5°C. What is the heat of combustion (energy released per gram) of the fuel?

Homework Equations

ΔU = Q - W
where ΔU = change of internal energy in system, Q = heat gained by system, W = work done on the surrounding BY the system.

The Attempt at a Solution

Now I know that for part a), heat has been transferred from the system, as the temperature of the water bath rises (meaning, heat has been transferred from the system to the surroundings). However, I am a little confused about part b). I am not sure if the work is done by the system or not. I am guessing, since the metal can is at constant-volume, no work has been done, but I may be wrong. Please correct me if I am wrong.
I am stuck at this point. I can't attempt the other parts of the problem unless I understand part b).



Thank you.

 

[voko]

Your understanding of b) is correct.

 

[NascentOxygen]

A student performs a combustion experiment by burning a mixture of fuel and oxygen in a constant-volume metal can surrounded by a water bath. During the experiment, the temperature of the water is observed to rise.

and

Now I know that for part a), heat has been transferred from the system, as the temperature of the water bath rises (meaning, heat has been transferred from the system to the surroundings). However, I am a little confused about part b). I am not sure if the work is done by the system or not.

The unit of work is the Joule. Work is done whenever energy is exchanged. Heat is measured in Joules. The chemical energy of the contents of the can is reduced. The heat energy of the water jacket increases. Work has been performed by the system on the water.

 

[cheahchungyin]

Imagine pushing a box.
Work is done ON the box, but BY who? Ya, BY you.
You are exerting kinetic energy towards the box to make it move :)
Imagine putting food into oven.
Work is going to be done ON the food (To cook it) BY the oven.
Note: There is only heat energy, no "cold" energy.
So when it comes to a fridge, heat energy is being transferred from the food to the fridge. Work done by the food :D

 

[Nickie]

I would like to clarify that the concept of work being "done by" or "on" the system can be confusing, and it is important to understand the difference between the two. In thermodynamics, work is defined as the energy transferred to or from a system due to a force acting on it. Therefore, work done "by" the system refers to the system doing work on its surroundings, while work done "on" the system refers to the surroundings doing work on the system.

In this particular experiment, the system is the mixture of fuel and oxygen, and the surroundings are the metal can and water bath. Since the metal can is at constant volume, no work is being done on or by the system. The only type of work in this scenario is pressure-volume work, which involves changes in volume, and since the volume is constant, there is no work being done.

Regarding part c), the sign of ΔU for the system would be negative, as heat is being transferred out of the system to the surroundings, resulting in a decrease in internal energy.

Finally, to solve part d), we can use the equation Q = mCΔT, where Q is the heat gained by the water, m is the mass of the water (200g), C is the specific heat capacity of water (4.186 J/g°C), and ΔT is the change in temperature (35.5°C). This will give us the total heat released by the fuel during combustion. To find the heat of combustion per gram of fuel, we can divide this value by the mass of fuel (1g) to get our final answer.