A 250-g sample of water at 20.0C is placed in a freezer that is held at a constant temperature of -20.0C. Considering the water as the “system,” answer the following questions:
a. What is the sign of qsys for the water after it is placed in the freezer?
b. After a few hours, what will be the state of the water?
c. How will the initial enthalpy for the water compare with the final enthalpy of the water after it has spent several hours in the freezer?
d. What will the temperature of the water be after several hours in the freezer?
-
solid if at equilibrium
Tinitial = 20 C; Tfinal = -20
-20 if at equilibrium
a. The sign of qsys for the water after it is placed in the freezer is negative. This is because heat is being removed from the water, resulting in a decrease in temperature, indicating an exothermic process.
b. After a few hours, the state of the water will be in a solid form, specifically ice. This is because the constant temperature of -20.0C in the freezer causes the water to freeze and undergo a phase change from a liquid to a solid.
c. The initial enthalpy for the water will be higher than the final enthalpy after it has spent several hours in the freezer. This is because enthalpy is a measure of the total heat content of a system, and since heat is being removed from the water, the final enthalpy will be lower than the initial enthalpy.
d. The temperature of the water after several hours in the freezer will also be -20.0C. This is because the freezer is held at a constant temperature of -20.0C, which means the water will reach thermal equilibrium with the freezer and match its temperature.
a. The sign of qsys for the water after it is placed in the freezer is negative (-). This is because the water is losing heat to the surroundings (freezer), resulting in a decrease in its internal energy.
To calculate the value of qsys, we can use the equation:
qsys = m * C * ΔT
where:
m is the mass of the water (250 g),
C is the specific heat capacity of water (4.18 J/g°C),
ΔT is the change in temperature (final temperature - initial temperature).
In this case, the final temperature is -20.0°C and the initial temperature is 20.0°C. Since the final temperature is lower than the initial temperature, ΔT will be negative. Plugging in the values:
qsys = (250 g) * (4.18 J/g°C) * (-20.0°C - 20.0°C)
= -20,900 J
The negative sign indicates that the water is losing 20,900 J of heat to the surroundings.
b. After a few hours in the freezer, the water will freeze and change its state from a liquid to a solid (ice). This is because the freezer is below the freezing point of water, causing the water molecules to slow down and form a solid structure.
c. The initial enthalpy for the water is the heat content at the initial temperature (20.0°C). The final enthalpy for the water after spending several hours in the freezer will be the heat content at the final temperature (-20.0°C).
Enthalpy (H) can be calculated using the equation:
H = m * C * T
where:
m is the mass of the water,
C is the specific heat capacity of water,
T is the temperature.
Since we are considering the water as the system, the enthalpy change (ΔH) is given by:
ΔH = m * C * ΔT
Both the initial and final enthalpies depend on the same mass (250 g) and specific heat capacity (4.18 J/g°C) of water. The only difference is the temperature.
As the final temperature (-20.0°C) is lower than the initial temperature (20.0°C), the final enthalpy will be lower than the initial enthalpy. This is because the water has lost heat and its internal energy has decreased.
d. The temperature of the water after several hours in the freezer will be the same as the temperature of the freezer, -20.0°C. This is because the water will reach thermal equilibrium with the freezer, where the ice and the freezer are at the same temperature.