Question

How much energy is absorbed when 10g of ice at 0 degrees becomes steam at 100 degrees please help

Answer 1

There are three heats to consider:

q1 = Heat required to melt the ice to water at 0°C.

q2 = Heat required to warm the water from 0 °C to 100 °C.

q3 = Heat required to vapourize the water to vapour at 100 °C.

q1 = mΔHfus
     = 10.g×334J1g 
     = 3340 J

q2 = mcΔT
      =10.g×4.184J1°C∙g×100.00°C 
      = 4180 J

q3 = mΔHvap
     =10.g×2260J1g
     = 22600 J

q1+q2+q3 = 3340 J + 4180 J + 22600 J
                  = 30120 J
                  = 30.12 kJ

∴ Converting 10 g of ice at 0 °C to steam at 100°C requires 30.12 kJ of energy.

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