One gram of water is placed in a cylinder and the pressure is maintained at 2.0 x 105 Pa. The temperature of the water is raised by 31oC. Determine the change in internal energy when, a) the water is in its liquid phase and causes an expansion of 1.0 x 10-8 m3. b) Water is in its gaseous phase and expands by the much greater amount of 7.1 x 10-5 m3. Assume that there is no phase change for both cases. (cliquidH20 =4186 J/(kgoC); cgasH20 = 2020 J/(kgoC))
To calculate the change in internal energy, we need to use the formula:
ΔU = ΔQ - ΔW
where ΔU is the change in internal energy, ΔQ is the heat added or removed from the system, and ΔW is the work done on or by the system.
a) When the water is in its liquid phase and causes an expansion of 1.0 x 10^(-8) m^3:
First, let's calculate the heat added or removed from the system (ΔQ).
ΔQ = m * c * ΔT
where m is the mass of water, c is the specific heat capacity of water (in its liquid phase), and ΔT is the change in temperature.
Given:
Mass of water (m) = 1 gram = 0.001 kg
Specific heat capacity of water (c) = 4186 J/(kg·°C)
Change in temperature (ΔT) = 31°C
ΔQ = (0.001 kg) * (4186 J/(kg·°C)) * (31°C) = 129.866 J
Next, let's calculate the work done on or by the system (ΔW).
ΔW = P * ΔV
where P is the pressure and ΔV is the change in volume.
Given:
Pressure (P) = 2.0 x 10^5 Pa
Change in volume (ΔV) = 1.0 x 10^(-8) m^3
ΔW = (2.0 x 10^5 Pa) * (1.0 x 10^(-8) m^3) = 2.0 J
Now, we can calculate the change in internal energy (ΔU):
ΔU = ΔQ - ΔW = 129.866 J - 2.0 J = 127.866 J
So, the change in internal energy when the water is in its liquid phase and causes an expansion of 1.0 x 10^(-8) m^3 is 127.866 J.
b) When the water is in its gaseous phase and expands by 7.1 x 10^(-5) m^3:
Using the same ΔQ and ΔW formula, the only difference is now we use the specific heat capacity of water in its gaseous phase (cgasH20 = 2020 J/(kgoC)).
Given:
Mass of water (m) = 1 gram = 0.001 kg
Specific heat capacity of water (c) in gaseous phase = 2020 J/(kg·°C)
Change in temperature (ΔT) = 31°C
Pressure (P) = 2.0 x 10^5 Pa
Change in volume (ΔV) = 7.1 x 10^(-5) m^3
Calculate ΔQ:
ΔQ = (0.001 kg) * (2020 J/(kg·°C)) * (31°C) = 62.62 J
Calculate ΔW:
ΔW = (2.0 x 10^5 Pa) * (7.1 x 10^(-5) m^3) = 14.2 J
Calculate ΔU:
ΔU = ΔQ - ΔW = 62.62 J - 14.2 J = 48.42 J
So, the change in internal energy when the water is in its gaseous phase and expands by 7.1 x 10^(-5) m^3 is 48.42 J.