A 25kg copper at 100c is placed in a container of water. the water has a mass of 45kg and it is at 45c beffore the copper was added. what is the final temperature of the mixture?
4 answers
yes
want to know how to solve this problem
[mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal - Tinitial)] = 0
Substitute the following from the porblem:
mass Cu = 25,000 g
specific heat Cu = Look this up in your text/notes/web and make sure the units are J/g*C.
Tfinal Cu = unknown. Solve for this.
Tinitial Cu = 100 celsius from the problem.
mass H2O = 45.000 g from the problem.
specific heat H2O = I remember this as 4.18 J/g*C
Tinitial H2O = 45 celsius from the problem.
Tfinal H2O = this is the unknown and is the same as Tfinal Cu. So I would substitute x for each of these and solve for x. Post your work if you get stuck.
Substitute the following from the porblem:
mass Cu = 25,000 g
specific heat Cu = Look this up in your text/notes/web and make sure the units are J/g*C.
Tfinal Cu = unknown. Solve for this.
Tinitial Cu = 100 celsius from the problem.
mass H2O = 45.000 g from the problem.
specific heat H2O = I remember this as 4.18 J/g*C
Tinitial H2O = 45 celsius from the problem.
Tfinal H2O = this is the unknown and is the same as Tfinal Cu. So I would substitute x for each of these and solve for x. Post your work if you get stuck.
heat lost by copper is gained by water
25 kg * (100 - t) * (s.h. Cu) = 45 kg * (t - 45) * (s.h. H2O)
s.h. Cu is ... 385 J/kg⋅ºC
s.h. H2O is ... 4184 J/kg⋅ºC
25 kg * (100 - t) * (s.h. Cu) = 45 kg * (t - 45) * (s.h. H2O)
s.h. Cu is ... 385 J/kg⋅ºC
s.h. H2O is ... 4184 J/kg⋅ºC