posted by calous --- HELP PLEASEE!!! .
1 kg of ice at 0° C is mixed with 9 kg of water at 50° C (The latent heat of ice is 3.34x105 J/kg and the specific heat capacity of water is 4160 J/kg). What is the resulting temperature?
Heat into ice = 1*3.34*10^5 +(T-0)(4160)
Heat out of water = 4160 (50-T)(9)
set equal and solve for T
omg thank you so much!!
but ? where did the 1.3 come frm
1*3 is one times 3
1 kilogram times latent heat of ice
budhu hai kya tu