text: how much heat is absorbed by 1.00 x 10^2 g of ice at -20.0 degrees to become water at 0.0 degrees? in joules plz that works, but you need to add the latent heat of fusion required to make the ice into water.  you can do the e=smt where e is the energy required (in j), s is the specific heat of ice(2100 j/kg per degree c), m is the mass of ice (kg), and t is the change in temp (c).  e=(2100)(.1)(20) e=4200 j  then to make it into water, you need to do e=ml, there e is the energy required (j), m is the mass (kg), and l is the laternt heat of fusion (about 3.3 x 10^5 j/kg for water).  e=(.1)(3.3 x 10^5) e=33000 j  then add the two e's together to ge a total  e=37200 j