Recent content by Tafe

Thank you so much gneill. :)

The ice is now water, so ρwater = 1.0x103 kg/m3 ρ = m / V V = m / ρ V = (1x106 kg) / (1.0x103 kg/m3) V = 1000 m3

It makes so much more sense now! I think I got it. Heat required in phase change, ice at 0 °C to water at 0 °C: Q = mLf Q = (1x106 kg)(334x103 J/kg) Q = 3.34x1011 J Heat required in heating water at 0 °C to 1x10-20 °C: Q = mcΔT Q = (1x106 kg)(4190 J/kg*K)(1x10-20 °C - 0 °C) Q = 4.19x10-11 J...

Homework Statement A 1x106 kg piece of ice is placed into a lake. How much heat is taken from the lake to raise the temperature of the ice from 0 °C to 1x10-20 °C? How much volume does the lake increase by? Latent heat for water is 334x103 J/kg Homework Equations Found in my textbook, cice =...