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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)

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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.
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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  • $\begingroup$ 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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