How to determine enthalpy

Question

Insrant hot packs works by crystallizing sodium acetate (NaCH3COO). The molar enthalpy of crystallization for sodium is -56.7kj/mol. How many grams of sodium acetate are needed to warm 125.0mL of water from 21.0C to 35.4C

Answer 1

So, let's first figure out how much heat energy (enthalpy) is needed to raise the temperature of 125.0 ml of water from 21ºC to 35.4ºC. We can do this by using the formula q = mC∆T

q = heat = ?

m = mass = 125.0 ml x 1 g / ml = 125 g (we assume the density of water is 1.0 g / ml)

C = specific heat for water = 4.184 J/gº

∆T = change in temperature = 35.4º - 21.0º = 14.4º

q = (125 g)(4.184 J/gº)(14.4º) = 7531 J = 7.531 kJ

Now that we have that value, we can figure out how much sodium acetate we need to produce that much heat.

7.531 kJ x 1 mol NaCH₃COO / 56.7 kJ = 0.1328 mols NaCH₃COO needed

Now simply convert this to grams of NaCH₃COO by using the molar mass of 82.0 g / mol

0.1328 mols NaCH₃COO x 82.0 g / mol = 10.9 g NaCH₃COO (to 3 sig. figs.)