Global Warming and Expanding Ocean Volume

Algebra Level 3

Scientist who have studied historical climate data believe that the oceans are increasing in temperature faster now that at any point in the last 10000 years. Over the last 60 years, the oceans have increased in temperature by . 18 .18 degrees C.

While this doesn't sound like much of a change, the density of water changes as its termperature increases. For example, water at 17 C has a density of 998.8 998.8 kg/m 3 ^3 , but water at 18 C has a density of 998.6 998.6 kg/m 3 ^3 . Assuming that in the temperature range we are considering the density function of water is roughly linear, how many more cubic km of ocean water on Earth are there now than there were 60 years ago? (give the nearest answer)

Assume that the current volume of the ocean is approximately 1.333 × 1 0 9 km 3 . 1.333 \times 10^9 \text{ km}^3.

5 × 1 0 4 5 \times 10^4 5 × 1 0 0 5 \times 10^0 5 × 1 0 6 5 \times 10^6 5 × 1 0 2 5 \times 10^2

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Yathish Dhavala
Feb 11, 2014

Change in density/change in temperature = -0.2/1 = -0.2 Since density function is linear it should be of the form density, D = -0.2T + K ; where T is temperature To find constant K put initial values (998.8,17) or final values (998.6,18). This gives K = 1002.2 but this is not needed.

Anyway, density function is D = 1002.2 - 0.2T

Differentiate this: dD = -0.2 dT dT = 0.18 (over last 60 years)

Hence dD = -0.2*0.18 = -0.036

Now, density=mass/volume D=m/V differentiate this: dD/D = -dV/V

Hence Change in volume dV = - V dD/D = - 1.333*10^{9} * (-0.036)/ 1000 = 5 x 10^{4} (approximately)

D = 1000 kg/m^{3} is taken as density of water

Positive dV implies the volume is increasing...

Instead of so much calculus, given that the question is asking for an approximate answer, you can simply see int he percentage terms.

Data given in the question says 0.2 change occurs (on a base of about 1000) in density per degree (unit of 0.2 is immaterial because we are talking in percentage terms). Hence, for 0.18 degree change, expected change would be 0.036 on a base of 1000. This gives a change of 3.6 * 10^-5 times. Multiplying this by 1.333 * 10^9 gives an approximate figure of 4.8 * 10^4

Ishant Goyal - 7 years, 3 months ago

very nice.....thanx

Pathik Patel - 7 years, 4 months ago

current volume is in km^3 , density is given in kg/m^3 i did,nt see changing unit of current volume ... please explain

Hari krishnan - 7 years, 4 months ago

Log in to reply

If we convert V = 1.333x10^9 km^3 = 1.33 x 10^18 m^3 Hence we'll get dV = 5x10^13 m^3

Presumably answer is needed in km^3. So converting this dV back to km^3 will give the same 5x10^4 km^3

Essentially there is no need to change any units (Or you can if you like but the answer will be same)..

Yathish Dhavala - 7 years, 4 months ago

gr8 soln!!

Shikhar Jaiswal - 7 years, 3 months ago

Since the current temperature wasn't given why was it assumed that the current density of water is 1000kg/m^3?

Alexander Alcantara - 7 years, 3 months ago

how find value of k and dt?????

Lalita Verma - 7 years, 3 months ago
Rishabh Raj
Feb 17, 2014

Assuming,the change(decrease) in density is proportional to the change(increase) in volume, [the mass being constant].

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...