Question

Datasheet Table 1: Calorimeter Constant AThot qhot ATcold gcold cal Tcal Ccal T cold Thot final Mass of Hot water* Mass of Cold water* 50 S0 * 1 mL of water has a mass of 1 gram Table 2: Heat of Dissolution ATso initial Tfinal Mass of Water Mass of NaOH gsol 1009 ATcal qcal Mass of Solution104 qdis

Answer 1

For Table 1

T_cold = 20^oC = 293 K

T_hot = 63^oC = 336 K

T_final = 39^oC = 312 K

Mass of hot water = 50 g = 0.05 kg

Mass of cold water = 50 g = 0.05 kg

ΔT_hot = 336-312 = 24 K

q _hot = ?

There is a general equation, q = mcΔT

Where q = heat energy, m = mass, c = specific heat capacity, and ΔT = temperature change.

The specific heat of water is 1 calorie/gram °C =4186 joule/kg. K. So, putting the value in the equation,

q = 0.05 x 4186 x 24

q_hot = 5023.2 Joules = 5.02 kJ.

ΔT_cold = 312-293 = 19 K

Therefore, q = 0.05 x 4186 x 19

q_cold = 3976.7 Joules = 3.98 kJ.

For Table 2

T_initial = 20^oC = 293 K

T_final = 29^oC = 302 K

Mass of water = 100 g = 0.1 kg

Mass of NaOH = 4.12 g = 0.004 kg

Mass of solution = 104.12 g = 0.10412 kg

ΔT_sol = 302-293 = 9 K

q _sol = ?

The specific heat of NaOH-water solution is approximately 3772 joule/kg. K. Therefore,

q _sol = 0.10412 x 3772 x 9

q_sol = 3534.7 Joules = 3.53 kJ.