A steel cylinder has a volume of 0.26 L and holds 4 kg of water at 25˚C and 200 kPa. (a) Find the total mass and total volume of the system (cylinder + water). (b) Give the values for two EXTENSIVE properties and two INTENSIVE properties of the water.
To find the answers to the questions, we need to use the given information and apply relevant formulas and concepts. Let's go step by step:
(a) Total mass and total volume of the system (cylinder + water):
First, let's calculate the mass of water using its volume and density. The density of water at 25˚C is approximately 1 kg/L. Therefore, the mass of water is given by:
Mass of water = Volume of water × Density of water
Mass of water = 0.26 L × 1 kg/L
Mass of water = 0.26 kg
Now, we can find the total mass by adding the mass of water to the mass of the cylinder. The given mass of water is already 4 kg. So, the total mass is:
Total mass = Mass of water + Mass of cylinder
Total mass = 4 kg + 0.26 kg
Total mass = 4.26 kg
To find the total volume, we can simply use the given volume of the cylinder, which is 0.26 L.
Total volume = Volume of cylinder
Total volume = 0.26 L
Therefore, the total mass of the system is 4.26 kg and the total volume is 0.26 L.
(b) Extensive and intensive properties of water:
Extensive properties are those that depend on the amount of substance, whereas intensive properties remain the same regardless of the amount. Two examples of each are:
Extensive properties:
1. Mass: As we saw in part (a), the mass of water changes depending on the amount present.
2. Volume: The volume of water can vary depending on the container it is in or the amount of water.
Intensive properties:
1. Temperature: The temperature of water remains the same regardless of the amount present.
2. Pressure: The pressure of water remains the same regardless of the amount present. In this case, it is given as 200 kPa.
Therefore, the two extensive properties of water are mass and volume, while the two intensive properties are temperature and pressure.