Q=55034175 J
c(ice)=2060 J/kg•℃
λ=330000 J/kg
c(water)=4183 J/kg•℃
Q=Q₁+Q₂+Q₃
Q=c(ice)m(12.15-0)+λm+c(water)m(t-0),
Q/m=12.15•c(ice) +λ+c(water)•t.
Solve for “t”
If 55034.175 kJ of heat are transferred to 150 kg of ice at a temperature of -12.15 degrees C, calculate the temperature of the resulting water.
I know my formulas for sensible and latent heat but don't know how to approach this.
1 answer