A steel cylinder of mass 2 kg contains 4 L of liquid water at 25^∘ C at 200 kPa. Find the total

A steel cylinder of mass 2 kg contains 4 L of liquid water...

A steel cylinder of mass $2 \mathrm{kg}$ contains $4 \mathrm{L}$ of liquid water at $25^{\circ} \mathrm{C}$ at $200 \mathrm{kPa}$. Find the total mass and volume of the system. List two extensive and three intensive properties of the water.

Key Concepts

System Analysis in Thermodynamics

In thermodynamics, it is essential to carefully define the boundaries of the system that is being analyzed. By clearly identifying all components (for example, a container and its contents), one can correctly apply the concepts of extensive and intensive properties to determine overall system attributes like mass and volume.

Conservation of Mass

The conservation of mass principle states that the total mass of a closed system remains constant over time, regardless of the physical or chemical changes occurring within it. This concept is critical when calculating the total mass of a system by adding the masses of individual components, such as a steel container and the water it holds.

Extensive Properties

Extensive properties depend on the quantity of matter in a system and are additive when systems are combined. In thermodynamics, examples include mass and volume. These properties help in understanding how changes in the size or amount of material (like water mass or container material) affect the overall system.

Intensive Properties

Intensive properties do not depend on the system size or the amount of material present. Key examples include temperature, pressure, and density. These properties define the state of a substance (such as water) and remain the same whether the substance is divided or combined.

Transcript

00:01 So in this problem, we want to find out the total mass and volume of this system.

00:06 And the system here is a steel cylinder filled with four liters of water.

00:12 So the first thing that we're going to do is set up what we need to know in order to solve for the total mass and the volume.

00:19 So what the total volume is going to be is the volume of the steel, the cylinder that the water is kept in, plus the volume of the water itself.

00:31 And for the total mass, similarly, it's going to be the mass of the steel and the mass of the water.

00:42 So what we already know what was given to us is we know the volume of the water and we know the mass of the steel.

00:48 So now what we're looking for is the volume of steel and the mass of water.

00:51 So let's start with finding the volume of the steel.

00:56 So we can solve for this by rearranging the equation that we have for density.

01:04 Where the density is equal to the mass divided by the volume.

01:08 So if we rearrange, we'll end up getting this equation, where the volume is equal to the mass divided by the density.

01:17 So we have the density of steel here, 7 ,820 kilograms per meters cubed.

01:24 So when we plug in for the volumes that we know, or for the values that we know, we end up getting is a value of 0 .256.

01:36 Times 10 to the negative third meters cubed...

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