Question

A 100 gram piece of ice is placed in an insulated calorimeter of negligible heat capacity containing 100 grams of

water at 373K.

a. What is the final temperature of the water once thermal equilibrium is established?

b. Find the entropy change of the universe for this process.

Answer 1

a) 10.23°C

b) - 0.8795 kJ / k

Step-by-step explanation:

Given data :

mass of ice = 100 gram = 0.1 kg

mass of water in calorimeter = 100 grams = 0.1kg

temperature of water in calorimeter = 373K

a) Determine the Final temperature of the water once thermal equilibrium is established

Heat lost by water = heat gained by Ice

Mwater * CPwater * Δt = Mice * Lice + Mice * CPwater * Δt

0.1 * (4.2 * 10^3 ) * ( 100 - t ) = ( 0.1 * 334 * 10^3 ) + 0.1 * (4.2 *1063 ) * ( T - 0 )

42000 = 33400 + 0.2 * 4200 * T

∴ equilibrium temp ( T ) = 10.23°C

B) Determine the entropy change of the universe of the process

Entropy change of ice = 33400/ 273 + 0.1 * 42 * 10^3 * In 283.23 / 273

= 276.84 J/K

Entropy change of water = 0.1 * 42* 10^3 * In 283.23 / 273

= -1156.34 J/K

Hence Entropy change of variance = 276.84 J/K + ( - 1156.34 J/K )

= - 0.8795 kJ / k