Question

Typical residential shower controls mix streams of hot water (140 F, or 60 C) with cold water (60 F, or 15 C) to form a stream of 40 C (104 F) water. The entire system loses energy to the surroundings at a rate of 5 kJ/kg of exiting water. What is the ratio of cold water-to-hot water mass flow rates necessary to provide the 40 C water?

Answer 1

Answer:

  4:5

Explanation:

Let x represent the fraction of the mix that is hot water. Then the temperature of the mix is ...

  60x +15(1-x) = 40·1

  45x = 25 . . . . . . . . . subtract 15

  x = 25/45 = 5/9 . . . divide by the coefficient of x

This is the fraction that is hot water, so the fraction that is cold water is ...

  1-5/9 = 4/9

The ratio of cold to hot is ...

  cold : hot = (4/9) : (5/9) = 4 : 5

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Additional comments

The problem assumes that the energy contained in a given mass of water is proportional to its temperature. That is almost true, sufficiently so that we can reasonably use that approximation.

If heat loss is figured into the problem, then additional information is needed regarding the energy content of water at temperatures in the range of interest. That is not provided by this problem statement, so we have ignored the heat loss.