What is the total entropy change for a 0.350 kg piece of ice at an initial temperature of 0.00 degC if it is placed in a Styrofoam cup having 2.22 kg of water at an initial temperature of 22.9 degC.

Answer 1

dQ = T ds

Where dQ= heat given or taken

T temperature = 273+0 = 273 K

dS = Change in entropy (J/K)

In this case

dQ = Mass*Sp heat* Change in temp

Mass = 0.35 Kg

Sp heat = 4200 J/kg K

dS = 0.35 *4200 * (22.9 - 0)*/ 273 J/K

Here Change in temp = 22.9K only

sp heat of ice at 0degree C is 4200 J/kg*K