What was the heat lost to the surroundings if #"11.3 g"# of diamond is cooled from #38.7^@ "C"# to #21.3^@ "C"#?
1 Answer
Jul 17, 2015
Well, if ever you're lucky enough to get this much diamond... the specific heat capacity of diamond is available here.
It is approximately
Thus, using the equation for heat transfer at constant pressure:
#q = mcDeltaT = mc(T_f - T_i)#
where:
#q# is heat flow in#"J"# .#m# is mass in#"g"# .#c# is specific heat capacity in#"J/g"^@ "C"# or#"J/g"cdot"K"# .#DeltaT# is the change in temperature in#""^@ "C"# or#"K"# .
So,
#q = ("11.3 g")("0.516 J/g"^@ "C")(21.3^@"C" - 38.7^@"C")#
#= -101.456 -> -"102 J"#
Since the temperature dropped, heat transferred out of diamond. To account for the sign...
#=# #"102 J lost"# to the surroundings.
