One gram of water changes from liquid to solid at a constant pressure of one atmosphere and a constant temperature of 0°C. In the process, the volume changes from 0.90 cm3 to 1.10 cm3.
The work done by the water is .02J, but I'm not sure how to find the internal energy. I tried 0, but that was wrong.
To find Work
Use the Work = Pressure * delta Volume
Be sure everything is in SI units
1atm=1.013*10^5
1cm^3=1*10^-5m^3
I have no idea how to find the change in internal energy though sorry
To find the change in internal energy, you need to use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
In this case, since the water changes from liquid to solid, it undergoes a phase change. During a phase change, the temperature remains constant. This means that the change in internal energy is solely due to the heat added or removed from the system.
Since the internal energy change (∆U) is equal to the heat added (Q) minus the work done (W), we can use the equation:
∆U = Q - W
In this case, the work done by the water is given as 0.02 J. However, the problem does not provide information about the amount of heat added or removed (Q).
To find the change in internal energy, you need to know the heat added or removed during the process. Without this information, we cannot determine the exact value of the change in internal energy (∆U).