Asked by stefani
How much energy in kJ is released when 17.55 g of water at 47.5 °C is cooled to –33.5 °C?
M.P. = 0 °C; B.P. = 100 °C; All specific heats in J/(g*K): s(ice) = 2.03; s(water) = 4.184; s(steam) = 1.890. heat of fusion = 6.01 kJ/mol; heat of vaporization = 40.67 kJ/mol.
M.P. = 0 °C; B.P. = 100 °C; All specific heats in J/(g*K): s(ice) = 2.03; s(water) = 4.184; s(steam) = 1.890. heat of fusion = 6.01 kJ/mol; heat of vaporization = 40.67 kJ/mol.
Answers
Answered by
DrBob222
q1 = heat released to cool from 47.5 to zero C.
q2 = heat released at 0 C to freeze water @ zero C.
q3 = heat released to cool from zero C to -33.5 C.
Total is q1 + q2 + q3
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2 = mass H2O x heat fusion.
q3 = mass ice x specific heat ice x (Tfinal-Tinitial)
Note: q1 will be in J if you use 4.184 for sp. heat.
q2 will be in kJ AND not it is in kJ/mol; therefore, you must change that to kJ/g or change mass H2O from grams to mols.
q3 will be in J if you use 2.03.
finally, q1, q2, q3 must be in J or kJ to add them.
Post your work if you get stuck.
q2 = heat released at 0 C to freeze water @ zero C.
q3 = heat released to cool from zero C to -33.5 C.
Total is q1 + q2 + q3
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q2 = mass H2O x heat fusion.
q3 = mass ice x specific heat ice x (Tfinal-Tinitial)
Note: q1 will be in J if you use 4.184 for sp. heat.
q2 will be in kJ AND not it is in kJ/mol; therefore, you must change that to kJ/g or change mass H2O from grams to mols.
q3 will be in J if you use 2.03.
finally, q1, q2, q3 must be in J or kJ to add them.
Post your work if you get stuck.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.