12. Water is generally said to be nearly incompressible. The deepest part of the

Question

12. Water is generally said to be nearly incompressible. The deepest part of the ocean abyss lies
at the bottom of the Marianas trench off the Philippines, at a depth of nearly eleven kilometers.
At a depth of 10.0 km, the measured water pressure is an incredible 103 MPa (that's
megapascals). (a) If the density of seawater is 1030 kg/m3 at the surface of the ocean, and its
bulk modulus is B = 2.34×109 N/m2, what is its density at a depth of 10.0 km? (Hint: Use the
volume stress equation from the study of elasticity, P = B (V/Vi), where the object
undergoing the stress has an initial volume Vi, and experiences a change in volume, V, when
the pressure changes by P.) (b) By what factor does the density increase?
(a)    kg/m3
(b)            ?

Explanation / Answer

Hey there,

Using the hinted formula P = B (V/Vi), we are using volumes. But of course, volume is just mass/density, so if we assume a singular unit of mass for both measured volumes (i.e. 1 kg), we are given:
B = 2.34x109 N/m2
P= 103MPa or 103x106 Pa
Vi = (1 Kg)/(1030 Kg/m3) = 9.7087x10-4 m3

So, by plugging in our givens, we acquire a V of -4.2735x10-5 m3, which means our new volume is just the old one plus the change, or (9.7087x10-4 - 4.2735x10-5) m3 = 9.2813x104 m3. Finally, work the 1 Kg we assumed back in to get our new density, D = 1/9.2813x104 1077.

I'm assuming factor means relating the new density back to the old. Our new density is 1077, our old is 1030. Thus, D/Di = 1.046; which means the density increases by a factor of 1.046.

Hope this helps.

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.