Chemistry thermonuclear question

Question

The molar heat of a solution, ʌH soln , of NaOH is -44.51 kJ/mol . In a certain experiment, 36.5 grams of NaOH is completely dissolved in 100 mL of water at 22.0° C in a foam cup calorimeter. Calculate the final temperature of the water.
Please explain this to me, I have finals tomorrow and I'm far behind

Answer 1

This is a calorimetry problem. The heat generated by the NaOH dissolving will heat up both the water and the NaOH dissolved in it. The equation for calculating heat is q = m Cp ΔT, mass times heat capacity times change in Temperature. The heat lost by the NaOH is gained by the solution. Absent other information, we will assume that the density of water is 1.00 g/mL and the heat capacity of the NaOH solution is the same as that of water (4.184 J/g°C)

 

First, calculate the heat given off by dissolving, which requires you to convert the grams of NaOH to moles (the molar mass of NaOH is 39.998g/mol)

 

q lost = 36.5 g NaOH |      mol NaOH    | -44.51 kJ  | 10³ J     =    -40617.4 J

                               | 39.998 g NaOH |  mol NaOH | kJ

 

q gained by water = -q lost = - (-40617.4 J) = 40617.4 J

 

q gained by water = mass (C_p) (T_final - T_initial)       The mass of the solution is 100 mL (1.00g/mL) + 36.5 g = 136.5 g

 

40617.4 J = 136.5 g (4.184 J/g°C) (T - 22.0)°C

 

40617.4 J = 571.116  J/°C (T)  -  12564.5 J    (Put both numerical terms on the same side)

 

(40617.4 + 12564.5) J  = 571.116 J/°C  (T)

 

53181.9 J =  571.116 J/°C (T)           (divide both sides by 571.116 J/°C)

 

T  =  53181.9 J                  T  =  93.1 °C   (round at the end to 3 significant figures)

         571.116 J/g°C