Question

How much energy is required to change a 40 g ice cube from ice at -10 oC to steam at 110 oC?

cwater=4186 J kg-1 oC-1

cice=2090 J kg-1 oC-1

cvapor=2010 J kg-1 oC-1

Lf=3.3x105 J kg-1

Lv=2.26x106 J kg-1

Answer 1

Useful information:
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C

Step 1: Heat required to raise the temperature of ice from -10 °C to 0 °C Use the formula

q1 = mcΔT = 40* 2.09* ( 0--10) =836 J

Step 2: Heat required to convert 0 °C ice to 0 °C water

q2 = m·ΔH = 40 *334 = 13360 J

Step 3: Heat required to raise the temperature of 0 °C water to 100 °C water

q3 = mcΔT = 40 *4.18* ( 100-0) =  16720J

Step 4: Heat required to convert 100 °C water to 100 °C steam

q4 = m·ΔH_v = 40*2275 = 91000 J

Step 5: Heat required to convert 100 °C steam to 110 °C steam

q5 = mcΔT = 40*2.09*( 110-100) = 836 J

Step 6: Find total heat energy

Heat_Total = Heat_{Step 1} + Heat_{Step 2} + Heat_{Step 3} + Heat_{Step 4} + Heat_{Step 5}

= 836+13360+16720+91000+836

  =122752 J