Question

A 102 g piece of ice at 0.0°C is placed in an insulated calorimeter of negligible heat capacity containing 100 g of water at 100°C. Find the entropy change of the universe for this process?

135 J/K 134 J/K is wrong,

Answer 1

let

m_Ice = 102 g = 0.102 kg
m_water = 100 g = 0.1 kg

let T is the final equilibrium temperature.

Heat lost by water = Heat gained by Ice

m_water*C_water*(100 - T) = m_Ice*Lf + m_Ice*C_water*(T - 0)

0.1*4186*(100 - T) = 0.1*3.33*10^5 + 0.1*4186*T

==> T = 10.2 degrees celcius

change in etntropy = m_Ice*Lf/(0 + 273) + m_Ice*C_water*((10.2+273)/273) + m_water*C_water*ln((10.2 + 273)/373)

= 0.1*3.33*10^5/273 + 0.1*4186*ln(283.2/273) + 0.102*4186*ln(283.2/373)

= 19.7 J/K <<<<<<<<------Answer