Question

Lesson 10 Part B KCA & KCQ

Section II KCA: Experimental Exploration of Entropy Change in Mixing

We will first determine the entropy change in mixing heated solid with water, according to our calorimetry experiment, then the total entropy change in the phase change of ice to steam, and finally in the dissolving salt in water.

Entropy Change in Mixing Heated Solid with Water

We will use the data we obtained in the calorimeter experiment to determine the entropy change in mixing the heated solid with the water in the calorimeter.

Measurements

Mass of empty Calorimeter cup, m_c= 47.68 g Initial temperature of calorimeter cup T_ci = 24°C
Mass of water in cup, m_w= 95.41 g                 Initial temperature of water in calorimeter cup, T_wi= 24°C
Mass of metal, m_m = 56 g                                 Initial Temperature of metal, T_mi = 100°C
Final (equilibrium) temperature of mixture of water + metal in calorimeter cup, T_eq = 28°C
Specific heat capacity of water: c_w= 4.186 J/g-K
Specific heat capacity of metal c_m= 0.317 J/g-K
Specific heat capacity of aluminum: c_c = 0.90 J/g-K

Calculations

(1) Change in entropy of heated metal when mixed with water in calorimeter cup, at low temperature: ?S_m= ?Q_m/T_m-avg, where ?Q_m= m_mc_m(T_eq– T_mi) and T_m-avg= ½ (T_eq+ T_mi)

(2) Change in entropy of water in calorimeter when mixed with heated metal, at low temperature: ?S_w+c= ?Q_w+c/T_(w+c)-avg, where ?Q_w+c= (m_wc_w+ m_cc_c)(T_eq– T_(w+c)i) and T_(w+c)-avg= ½ (T_eq+ T_(w+c)i)

(3) The change in the entropy of mixing the metal with the water in calorimeter cup is ?S = ?S_m+ ?S_w

Entropy Change in the Change of Ice to Steam

We will use the data from phase change experiment to determine the entropy change in melting ice to water and heating the water to steam.

Measurements

Mass of ice = 125.52 g       Temperature of ice T_ice = 0°C
Mass of water from melted ice = 115.76 g            Temperature of water melted from ice Tw = 0°C
Temperature of steam T_steam= 100°C
Latent heat of fusion of ice: L_f = 334 J/g
Latent heat of vaporization: L_v= 2260 J/g

Calculations

(1) Change in entropy of the melting of the ice at constant T: ?S₁= Q₁/T_avg= m_iceL_f/T_avg, where T_avgis the average temperature at the energy supplied

(2) Change in entropy of the melting of the ice at constant: ?S₂= Q₂/T_avg= m_wc_w?T/T_avg, where T_avgis the average of the temperature difference T_avg= ½ (T_f+ T_i)

(3) Change in entropy of the melting of the ice at constant T: ?S₃= Q₃/T_avg= m_wL_v/T_avg, where T_avgis the average temperature at the energy supplied

(4) The total change in entropy in the change of ice to steam is given by ?S_tot= ?S₁+ ?S₂+ ?S₃

Answer 1