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?
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
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
#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?
