How many kilojoules are required to convert 115.0 g of ice at 0.0 ∘c to liquid water at 32 ∘c? the heat of fusion of water is 334 j/g, and the heat capacity of water is 4.184 j/g ∘c?

The answer is 53.8 kJ.
Solution:There are two major steps in converting ice to liquid water. It begins with a phase change when ice melts at 0.0°C, and then a temperature change when the liquid water rises in temperature from zero to 32°C.
The amount of heat involved with the phase change melting is given by
     q = (mass of water) (ΔHfus)
        = (115.0 g)(334 J/g) 
        = 38410 J = 38.41 kJ
The amount of heat involved with temperature change is 
     q = mcΔT
        = (115.0g)(4.184J/g°C)(32°C - 0.0°C)
        = 15397.12 J = 15.39712 kJ
Summing up the two values gives the total heat required to convert ice to liquid water:
     q = 38.41 kJ + 15.39712 kJ= 53.8 kJ

Mgo2 is the oxidation state of manganese oxide