How much energy is needed to change 44 grams of ice at -8° to steam at 107°?
1 answer:
Q=m(c∆t +heat of fusion + heat of evaporation)
m= 44g
c= 4.186 J/g.C
∆t= 107-(-8) =115 C
heat of fusion= 333.55 J/g
heat of evaporation=2260 J/g
Q=44(4.186*115 + 333.55 + 2260)
Q= 135297.36 J
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