# Paradox in Basic Heat Transfer?

**URL:** <https://boards.straightdope.com/t/paradox-in-basic-heat-transfer/694289>\
**Category:** Factual Questions\
**Created:** [July 29, 2014, 12:55am UTC](https://boards.straightdope.com/t/paradox-in-basic-heat-transfer/694289 "2014-07-29T00:55:09Z")\
**Posts on this page:** 5\
**Page:** 1

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**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\
**Post date:** [July 29, 2014, 12:55am UTC](https://boards.straightdope.com/t/paradox-in-basic-heat-transfer/694289/1 "2014-07-29T00:55:09Z")

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Suppose I want to remove 800W of heat with a fan. As I increase cfm supplied by the fan, the cooling effect should increase, true? So, dT should increase. This seems straight-forward enough. Yet, using Q=mcdT and holding Q constant, algebra says as m increases, dT decreases. However, this is counterintuitive. Surely, I am overlooking something. Can someone help me see where my logic is going astray?

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**Author:** ![beowulff](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/beowulff/32/542_2.png) [@beowulff](https://boards.straightdope.com/u/beowulff)\
**Post date:** [July 29, 2014, 1:49am UTC](https://boards.straightdope.com/t/paradox-in-basic-heat-transfer/694289/2 "2014-07-29T01:49:11Z")

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I don’t think you can use that equation for forced-air cooling.

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**Author:** ![Ludovic](https://avatars.discourse-cdn.com/v4/letter/l/7ab992/32.png) [@Ludovic](https://boards.straightdope.com/u/Ludovic)\
**Post date:** [July 29, 2014, 2:03am UTC](https://boards.straightdope.com/t/paradox-in-basic-heat-transfer/694289/3 "2014-07-29T02:03:39Z")

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Plus dT would increase in the object you are cooling, but not in the air you are using to cool the object. If you use a greater amount of air to cool the object the same temperature, then the dT _for the air_ would decrease.

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**Author:** ![Isilder](https://avatars.discourse-cdn.com/v4/letter/i/cc9497/32.png) [@Isilder](https://boards.straightdope.com/u/Isilder)\
**Post date:** [July 29, 2014, 2:54am UTC](https://boards.straightdope.com/t/paradox-in-basic-heat-transfer/694289/4 "2014-07-29T02:54:51Z")

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Q is not a constant.

You’d better understand what the constant is… c ? What is specific heat ?

Joules per degree per gram  
Well you’ve got the equation right there.

Potential energy in object in the form of heat is its c X mass X delta T.

Note that with no time variable in this equation you cannot do calculus to get power (energy flow rate.) or nor time taken to reach equilibrium directly from this.

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**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\
**Post date:** [July 29, 2014, 3:36am UTC](https://boards.straightdope.com/t/paradox-in-basic-heat-transfer/694289/5 "2014-07-29T03:36:38Z")

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> [@Isilder](#):
>
> Q is not a constant.
> 
> You’d better understand what the constant is… c ? What is specific heat ?
> 
> Joules per degree per gram  
> Well you’ve got the equation right there.
> 
> Potential energy in object in the form of heat is its c X mass X delta T.
> 
> Note that with no time variable in this equation you cannot do calculus to get power (energy flow rate.) or nor time taken to reach equilibrium directly from this.

Actually, to clarify the meaning of “m”… “m” is understood to be m-dot, or lbs/hr just a conversion of cfm (cubic feet per minute). Thus, there is time in the equation.