How much energy is required to change a 40.0-g ice cube from ice at ?10 oC to steam at 110 oC?

Answer:

1. Temperature change from ice to water:

      Q = mc_iceΔT

      ΔT =-10^οC to 0^οC =10^οC

      c = 2090J/kg^οC

      Q =(.04kg)(2090J/kg^οC)(10^οC) = 836Joules

2. Phase change from ice to water:

      Q = mL

      L =3.33x10⁵J/kg

      Q =(.04kg)(3.33x10⁵J/kg) = 13320Joules

3. Temp change from water to steam:

     Q =mc_waterΔT

      ΔT =0^οC to 100^οC =100^οC

      c = 4190J/kg^οC

      Q =(.04kg)(4190J/kg^οC)(100^οC) =16760Joules

4. Phase change from water to steam:

      Q = mL

      L = 2.26 x10⁶J/kg

      Q = (.04kg)(2.26 x10⁶J/kg) = 90400Joules

5. Temp change of steam to final temp of steam:

      Q =mc_steamΔT

      ΔT =100^οC to 110^οC

      c = 2010J/kg^οC

      Q =(.04kg)(2010J/kg^οC)(10^οC) =804Joules

Now add all numbers: 836J + 13320J + 16760J + 90400J +804J

                                                =122120 Joules