Answer to Question #121978 in General Chemistry for megan

Question

Using a heating curve, how many kJ of energy are needed for a 22 g ice cube at -25°C to be turned into steam at 150°C?

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

q = mcΔT

where

q = heat energy

m = mass

c = specific heat

ΔT = change in temperature

q = (22g)x(2.09 J/g·°C)[(0 °C - -25 °C)]

q = 1149.5J

Heat required to raise the temperature of ice from --25 °C to 0 °C = 1149.5 J

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

Use the formula for heat:

q = m·ΔH_f

where

q = heat energy

m = mass

ΔH_f = heat of fusion

q = (22 g)x(334 J/g)

q = 7348 J

Heat required to convert 0 °C ice to 0 °C water = 7348 J

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

q = mcΔT

q = (22 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]

q = 9196J

Heat required to raise the temperature of 0 °C water to 100 °C water = 9196 J

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

q = m·ΔH_v

where

q = heat energy

m = mass

ΔH_v = heat of vaporization

q = (22g)x(2257 J/g)

q =49654 J

Heat required to convert 100 °C water to 100 °C steam = 49654J

Step 5: Heat required to convert 100 °C steam to 150 °C steam

q = mcΔT

q = (22g)x(2.09 J/g·°C)[(150 °C - 100 °C)]

q = (22 g)x(2.09 J/g·°C)x(50 °C)

q = 2299 J

Heat required to convert 100 °C steam to 150 °C steam = 2299J

Step 6: Find total heat energy

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

Heat_Total = 1149.5J + 7348 J + 9196 J + 49654 J + 2299 J

Heat_Total = 69646.5J

Answer:

The heat required to convert 25 grams of -25°C ice into 150 °C steam is 69646.5 J or 69.65kJ