Question

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.

Answer 1

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.