Is the time to (naturally) decrease temperature of water by 5 degrees the same, regardless of the initial temperature? Imagine 3 glasses of water (a, b c) in a room temperature of 70F.

There are 3 scenarios:

a) initial water temperature: 120F. final temperature: 115F
b) initial water temperature: 110F. final temperature: 105F
c) initial water temperature: 100F. final temperature: 95F

The question is - does the time for water to decrease its temperature by 5F for all cases above (a,b,c) is the same? Or the time is different for each case?

  • $\begingroup$ related:ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html Newton's Law of Cooling $\endgroup$ – user81619 Nov 17 '15 at 0:03
  • $\begingroup$ So like ambient temperature is not factor? -1 $\endgroup$ – paparazzo Nov 17 '15 at 0:07
  • $\begingroup$ I guess the time would be the same then? $\endgroup$ – Tom Nov 17 '15 at 0:16
  • $\begingroup$ @Tom: no, the fact that the water is cooling means there is a colder body involved. Heat then flows from the hotter to the cooler object. The larger the temperature difference, the faster the cooling. See Newton's Law of cooling: en.wikipedia.org/wiki/Newton%27s_law_of_cooling $\endgroup$ – Gert Nov 17 '15 at 0:20
  • $\begingroup$ I think I get it. So for 70F room temperature, the time to decrease water temperature for above scenarios is: a (fastest time) and c (slowest time)? $\endgroup$ – Tom Nov 17 '15 at 0:23

No, they will cool at different rates

Newton's law of cooling states that

$$ \frac{dQ}{dt} \propto T-T_{env} $$ where $T$ is the temperature of the water and $T_{env}$ is the temperature of the environment. As you can see, the larger the temperature difference, the more quickly the water will lose heat. We can assume that the specific heat of water will not change much with temperature, and no phase changes occur. Therefore, hot water will decrease in temperature more quickly than warm water.


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