First i gotta say that i couldnt find out the result just with the given values (and also i dont believe it can be calculated
,thus we re gonna make some necessary assumptions:
*the solution volume is exactly 150 ml
*The NaCl(aq) is sufficiently dilute that its 'density' and 'specific heat' are about the same as those values for pure water: 1.00 g/mL and 4.18 J/(g*C), respectively.
*The system is completely isolated:No heat escapes from the calorimeter.
*The heat required to warm any part of the calorimeter other than the NaCl(aq) is negligible.
The heat produced by the reaction and retained in the calorimeter is
q= mass * specific heat * (delta)T
q = 150 mL * 1.00 g/mL * 4.18 J/(g*C) * (26-23)C = 1881 J