How do you calculate the change in entropy due to heat flow at a constant temperature?

Suppose a heat source generates heat at a rate of 60.0 W (1 W = 1 J/s). How much entropy does this produce per hour in the surroundings at 28.1 °C? Assume the heat transfer is reversible.

I got my answer to be 0.1993 entropy generated in one second but I need to find the amount of entropy in one hour. How can I do this last step?

1 Answer
Nov 19, 2017

Well, if you aren't sure how to do a unit conversion, I may as well check the rest of your answer.


The change in entropy for a reversible process could be given by:

#DeltaS = (q_(rev))/T#

where #q_(rev)# is the heat flow involved in #"J"# and #T# is the temperature in #"K"#. Reversible simply means it is infinitesimally slow.

So, the change in entropy is:

#DeltaS = ("60.0 J"/"s")/(28.1 + "273.15 K") = "0.199 J/s"cdot"K"#

In terms of #"J/hr"#:

#color(blue)(DeltaS) = "0.199 J"/cancel"s" xx (60 cancel"s")/cancel"min" xx (60 cancel"min")/"hr" = color(blue)("717 J/hr"cdot"K")#

Does it make physical sense that more joules of energy were produced in a larger amount of time?