Question

A container holding 0.25kg of steam at 140 degree Celsius, is placed in a freezer. How much energy is requires to get that steam to 140 degrees Celsius to ice of -25 c.

L(F) H20=332J/g

L(V)H20=2260J/g

C steam=1.996J/gc

C water=4.18J/gc

Cice=2.1J/g

Answer 1

$Q=m(L_F+L_V+40c_s+100c_w+25c_i)=786~kJ.$