Thermodynamic entropy — UPSC Physics Optional CSE 2013 Paper I

In Leh, temperature of ice on a cold winter night is measured as -20^\circ\mathrm{C}. Calculate the change in entropy when 1\ \mathrm{kg} of ice is converted into steam at 100^\circ\mathrm{C}. Given specific heat capacity of ice is 500\ \mathrm{cal\,kg^{-1}\,K^{-1}}, latent heat of ice is 3.36\times10^5\ \mathrm{J\,kg^{-1}}, latent heat of steam is 2.26\times10^6\ \mathrm{J\,kg^{-1}} and J=4.2\ \mathrm{J\,cal^{-1}}.