Question

The enthalpy of solution (dissolving) of sodium hydroxide is given below. Determine the change in temperature of a coffee cup calorimeter containing 150 ml of water when 5.00 g NaOH (40.00 g/mol) is added to the container. You may assume that the solution has the same specific heat and density as water.
NaOH(s) → NaOH(aq) ΔH =-44.51 KJ

a. +8.87°C

b. +2.70 C

c. 2.70°C

d. 8.87°C

Answer 1

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

$M=d\times V=1.00 g/mL\times 150 ml =150.0 g$

Mass of solution = m = M = 150.0 g

$NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol$

Moles of NaOH = $\frac{5.00 g}{40 g/mol}=0.125 mol$

Energy released when 0.125 moles of NaOH added in water = Q

$Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J$

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

$Q'= 5,563.8 J$

Specific heat of solution = c = 4.184 J/g°C

Change in temperature of the solution = $\Delta T$

$Q'=mc\times \Delta T$

$5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T$

$\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC$

The change in temperature of a coffee cup calorimeter is 8.87°C.