Question

A 10.1 g sample of NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. The temperature increases from 23.0 °C to ________°C. Specific heat of liquid water is 4.18 J/g-K and ΔH for the dissolution of sodium hydroxide in water is 44.4 kJ/mol.

Answer 1

Answer:

33.3 °C

Explanation:

You have two heat flows in this experiment.

Heat from solution of NaOH + heat to warm water = 0

                     q1                       +               q2               = 0

                   nΔH                     +             mCΔT           = 0

Data:

m(NaOH) = 10.1 g

            ΔH = -44.4 kJ/mol

 m(H2O) = 250.0 g

              C = 4.18 J/(K·mol)

            Ti = 23.0 °C

Calculation:

n = 10.1 g NaOH × (1 mol NaOH/40.00g NaOH = 0.2525 mol NaOH

q1 = 0.2525 mol × (-44 400 J/mol) = -11 210 J

m(solution) = m(NaOH) + m(water) = 10.1 + 250.0 = 260.1g

q2 = 260.1 × 4.18 × ΔT = 1087ΔT J

-11 210 + 1087ΔT = 0

                   1087ΔT = 11 210

                             ΔT = 11 210/108745 = 10.31 °C

ΔT = T2 - T1 = T2 - 23.0 = 10.73

                                      T2 = 23.0 + 10.31 = 33.3 °C

The temperature increases to 33.3 °C.

Answer 2

Hey there!:

moles of NaOH = 10.1 / 40 = 0.2525

heat  = ΔH   x  moles

= 44.4 x 0.2525

= 11.21 kJ

total mass  = 10.1 + 250 = 260.1 g

Q  = m Cp dT

11211    = 260.1 x  4.18 x dT

dT  = 10.3

T2  = 10.3 + 23  = 33.3 °C

temperature = 33.3 ºC°

Hope this helps!