Question

How much heat is required to convert 12.0 g of ice at-10.0 degrees C to steam at100.0 degrees C ? Express your answer in joules and in calories.

Answer 1

a)

The amount of heat required to raise the temparature of ice from -10 deg C to 0 deg C

        Q1 = mc ?T

              = (12.0x10^-3 kg)(2100 J/kg.oC)(10oC)

              = 252J

The amount of energy required to complitely melt into ice

         Q2 = m L

               = (12x 10^-3 kg)(333x10^3 J/kg)

               = 3996 J

The amount of heat required to raise the temparature of water from 0 oC to 100o C

         Q3 = mc ?T

                = (12.0x10^-3 kg)(4186 J/kg.oC)(100oC)

                 = 5023.2 J

The amount of energy required to complitely boiled the water

         Q3 = m L

               = (12x 10^-3 kg)(2260x10^3 J/kg)

               = 27120 J

So total heat is

           Q = Q1+ Q2+ Q3 + Q4

                = 36391.2 J

in calories = 1 cal = 4.2 j

so in calorie = 8664.57 cal