Why does this question make this simplifying assumption?

[laser]

Homework Statement

See description

Relevant Equations

A = Q/It, where A is area, Q is heat, I is intensity, t is time

The question says to "neglect the change of volume of the water". This confuses me. Why do we neglect the change of the water's volume? Let's say we didn't. A = Q/It, where A is area, Q is heat, I is intensity, t is time. Q = mc(deltatheta), mass is constant, specific heat is constant (as mass is constant) and change in temperature is constant. I and t are also constant.

The change of volume of water would change the density of water, yes, but the mass will remain constant.

Why does the question say to neglect the change of volume of the water here?

 

[kuruman]

$\dots$ but the mass will remain constant.

Does the volume of the container holding the water expand at the same rate as the water?

 

[laser]

Does the volume of the container holding the water expand at the same rate as the water?

Oh, what you are saying is that since the volume is expanding, the area will also expand?

 

[kuruman]

That's not what I'm saying although it is true. You have 200 litres of water in a 200-litre container. Say the heated water expands from 200 litres to 205 litres. For the mass to stay the same, the volume of the container must also expand to at least 205 litres. If it doesn't and there is no safety valve to let water out, the container will burst from the pressure because water is incompressible.

 

[laser]

That's not what I'm saying although it is true. You have 200 litres of water in a 200-litre container. Say the heated water expands from 200 litres to 205 litres. For the mass to stay the same, the volume of the container must also expand to at least 205 litres. If it doesn't and there is no safety valve to let water out, the container will burst from the pressure because water is incompressible.

Ah I see. Let's assume that the volume of the container also expands. Why do we have to assume that the water doesn't expand then?

 

[pasmith]

Are you doing work on the water to raise its temperature, or to decrease its density (since the mass is unchanged but the volume has increased)?

 

[kuruman]

Ah I see. Let's assume that the volume of the container also expands. Why do we have to assume that the water doesn't expand then?

The water and the container expand with different expansion coefficients. For example, aluminium has a volume expansion coefficient that is about 1/3 that of water. The author of the problem does not want you to get tangled up with expansion calculations for a mere [4 marks], hence the generous suggestion to "neglect the change of the volume of the water here." That would keep the original mass of the water the same and the calculation simple.

 

[Steve4Physics]

Why does the question say to neglect the change of volume of the water here?

From the supplied data, we can see that the final answer needs a precision of only 2 significant figures.

Note that the volume expansion coefficient of water is ~0.0002 /ºC. So for a 20ºC rise, the volume would increase by ~0.4%. So it hardly matters here.

I'd guess that the very considerate author of the question didn’t want students wasting their time worrying about this small variation. (Though based on my past experience, very few would have!)

(To find the mass of water,you could use the density at the mid-temperature (30ºC). But it hardly matters in this context.)

 

[PeroK]

Why does the question say to neglect the change of volume of the water here?

The volume of a fixed mass of water increases with temperature. We have 200 litres of water being raised from 20 degrees to 40 degrees, which will entail an increase in volume. The assumption may be there to fend off those of an excessively pedantic disposition, who might otherwise be scornful of the problem statement.