The density of liquid water is ρ=1008-T / 2 [kg / m^3] with T in ^∘ C. If the temperature incre

The density of liquid water is ρ=1008-T / 2 [kg / m^3] with...

The density of liquid water is $\rho=1008-T / 2$ $\left[\mathrm{kg} / \mathrm{m}^{3}\right]$ with $T$ in ${ }^{\circ} \mathrm{C}$. If the temperature increases $10^{\circ} \mathrm{C},$ how much deeper does a $1-\mathrm{m}$ layer of water become?

Key Concepts

Density

Density is defined as mass per unit volume and is a fundamental property of materials. In fluids, density can vary with temperature, affecting how the mass is distributed within a given volume and influencing other physical characteristics such as buoyancy and pressure.

Thermal Expansion

Thermal expansion refers to the change in a material’s dimensions—such as length, area, or volume—in response to a change in temperature. For liquids, increasing the temperature generally causes the molecules to move apart, resulting in a lower density and a larger volume for the same mass.

Conservation of Mass and Volume Change

When the total mass of a fluid is held constant, any decrease in density due to thermal expansion must be compensated by an increase in volume. In the context of a liquid column, this manifests as a deeper or taller column because the fluid expands to maintain its mass while its density decreases.

Transcript

00:01 So in this problem, we're looking to find how much this height will change of a vat of water, given that we increase its temperature by 10 degrees.

00:12 So delta h is what we're looking for as the final expression.

00:18 But what we're going to do first is we're going to solve for the change in density after we increase the water by 10 degrees.

00:27 So the change in density is simply the final density minus the initial.

00:32 And when we use this expression on the left and simplify, what we get for an expression of delta row is minus delta t over 2, which plugging in for delta t 10 degrees celsius will give us a value of minus 5 kilograms per meters cubed.

00:56 So because this is a linear expression here that relates the temperature to the density, i'm going to take the initial density here.

01:06 I'm going to assume that we're looking at a temperature of 25 degrees celsius before we increase the temperature, which will give us an initial density of 995 .5 kilograms per meter cubed.

01:33 So then what i'm going to do is if we look here at this, at this system where we have the height and the area.

01:43 This defines the volume.

01:45 So the volume can be written as the area times the height.

01:50 And we know the expression that relates density to the volume is mass divided by volume equal to density.

01:58 So i'm going to actually replace the expression v for volume with the area times the height.

02:04 So we'll get mass divided by 8 times h...

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