Add the entropy gains of the ice cube (as it heats up to the equilibrium T) and the cooling water (as is cools down), The original liquid water actually loses entropy, but the ice gains more.
You will need to calculate the equilibrium temperature first. All of the ice will melt in this case.
The water-cooling and ice-heating steps will require an integration of dT/T. The melting step will be isentropic.
The heating of the melted ice to the final equilibrium temperature also has an associated entropy gain that must be included.
a 9.0 g ice cube at -10 C is put into a thermos flask containing 100 cm^3 of water at 20 C. by how much has the entropy of the cube-water system changed when the equilibrium is reached? the specific heat of ice is 2220 J/kg K.
1 answer
1 answer