Question

15 grams of ice at -5° C are brought into thermal contact with 6 grams of steam at 100° C.

If you have C_ice = 2090 J/Kg°K , C_water = 4186 J/Kg°C , Lf = 3.33x10⁵ J/Kg  and Lv = 2.26x10⁶ J/Kg

Calculate the entropy change when the system reaches thermal equilibrium

Answer 1

mi = mass of ice = 15 g = 0.015 kg

msteam = mass of steam = 6 g = 0.006 kg

initial temperature of ice = -5 degree celsius = 268 K

initial temperature of steam = 100 degree celsius = 373 K

C_ice = 2090 J/Kg°K

C_water = 4186 J/Kg°C

Lf = 3.33x10⁵ J/Kg

Lv = 2.26x10⁶ J/Kg

Let the final temperature be T.

applying principle of calorimetry :

heat loss = heat gain

.

Total enetropy change is =

= 0.58 + 18.30 + 3.73 - 36.35 -6.34 = −20.08 J/K [answer]