Question

How much energy is required to convert 14.0 g of ice at −15 °C to 14.0 g of steam at 114°C?

Specific heat capacity, ice: 2.108 J/g·K (at 0°C)

Specific heat capacity, water: 4.1801 J/g·K (at 25°C)

Specific heat capacity, steam: 2.0784 J/g·K (at 100.°C)

ΔHfus = 333.6 J/g (at 0°C)

ΔHvap = 2.2567 kJ/g (at 100.°C)

Answer 1

Answer – Given, mass of water = 14.0 g , ti = -15.0^oC , tf = 114^oC

Specific heat capacity, ice: 2.108 J/g·K (at 0°C)

Specific heat capacity, water: 4.1801 J/g·K (at 25°C)

Specific heat capacity, steam: 2.0784 J/g·K (at 100.°C)

∆Hfus = 333.6 J/g (at 0°C)

∆Hvap = 2.2567 kJ/g (at 100.°C)

Calculate the heat from the -15^oC to 0.0^oC

We know heat formula

q1 = m*C solid * ∆t

       = 14.0 g * 2.108 J/g^oC * (0.0^oC –(-15^oC)

        = 442.68 J

Now heat from 0.0^oC to 0.00^oC

q2 = m*∆H fus

     = 14.0 g * 333.6 J/g

     = 4670.4 J

Heat from the 0.00^oC to 100^oC

q3 = m*C liq* ∆t

    = 14.0 g * 4.184 J/g^oC * (100.0^oC –0.0^oC)

     = 5857.6 J

heat from 100^oC to 100^oC

q4 = m*∆H vap

     = 14.0 g * 2256.7 J/g

     =31593.8 J

Heat from the   100^oC to 114^oC

q5 = m*C gas* ∆t

    = 14.0 g * 2.0784 J/ g^oC * (114^oC –100^oC)

     = 407.4 J

Total heat q = q1+ q2+q3+q4+q5

                         = 442.68 J + 4670.4 J + 5857.6 J + 31593.8 J + 407.4 J

                          = 42972 J

                        = 42.97 kJ

So, 42.97 kJ energy is required to convert 14.0 g of ice at −15 °C to 14.0 g of steam at 114°C