Assume that you have a wet shirt that you have just washed. You squeeze of the surplus water that can be squeezed out.

Method 1:

1)You iron the shirt for 5 minutes and then hang it out to dry out in the warm sunshine (morning sun) for 15 minutes and examine it.

Method 2:

2)You first hang it out in the sunshine for 15 minutes and then iron the shirt for 5 minutes (same heat setting) and then you examine it

Which of the above two methods is better for drying the shirt faster?

(If you need more data:

1)Assume cotton shirt 2)Assume that the wet shirt is not dripping water 3)Assume ambient temp is at 35 deg Celsius 4)Assume that it is not a windy or a cloudy day)


closed as off-topic by John Rennie, CuriousOne, user36790, John Duffield, Gert Mar 4 '16 at 0:17

  • This question does not appear to be about physics within the scope defined in the help center.
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  • 5
    $\begingroup$ I'm voting to close this question as off-topic because it isn't about physics $\endgroup$ – John Rennie Mar 3 '16 at 10:39
  • 2
    $\begingroup$ @JohnRennie If this is not a question on physics, this will not be one as well : physics.stackexchange.com/q/5265 $\endgroup$ – Sidarth Mar 3 '16 at 12:01
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    $\begingroup$ @JohnRennie: While I can't say I like this question, questions essentially about rates of evaporation seem to be technically physics to me. $\endgroup$ – ACuriousMind Mar 4 '16 at 0:14
  • $\begingroup$ This question definitely sounds like a homework, so it should remain closed, since there is no manifestation that you tried to understand the homework you proposed. $\endgroup$ – FraSchelle Nov 1 '16 at 8:53

Assuming iron to be pretty hot, ironing first and then drying in the sun will dry faster.

Suppose, 15 minutes in sun evaporates 50% of initial water content.

Suppose 5 minutes ironing evaporates 50% of initial water content in the wet shirt.

Suppose 5 minutes ironing evaporates 20% of initial water content in the already dried in sun shirt.

The reason is that iron being hot, will evaporate more % of water when the shirt is wet as compared to when it has already been dried in sun for 15 minutes. Evaporation % rate (for wet and dry shirt) would hardly differ for drying in sun.

Case 1 - Iron, then sun - 50% + 50% of remaining 50% = 75% dry.

Case 2 - Sun, then iron - 50% + 20% of remaining 50% = 60% dry.

  • $\begingroup$ This solution depends on the figures used in the assumptions. Change the figures and you can change the conclusion. There is no application of physics here, only mathematics. $\endgroup$ – sammy gerbil Oct 29 '16 at 23:39

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