Answer to Question #190223 in Civil and Environmental Engineering for John Prats

Question

Determine the amount of heat required to changed 200 g of ice at -5°C to steam at 105°C.

Answer 1

Required 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:

Find the heat required to raise the temperature of ice from -5 °C to 0 °C. Use the formula:

q = mcΔT

where

• q = heat energy
• m = mass
• c = specific heat
• ΔT = change in temperature

In this problem:

• q = ?
• m = 200 g
• c = (2.09 J/g·°C
• ΔT = 0 °C - -5 °C=5⁰ C
  

Plug in the values and solve for q:

q = (200 g)x(2.09 J/g·°C)[(5⁰ C]

q = (200 g)x(2.09 J/g·°C)x(5 °C)

q = 2090 J

The heat required to raise the temperature of ice from -5 °C to 0 °C = 2090 J

Step 2:

Find the 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

For this problem:

• q = ?
• m = 200 g
• ΔH_f = 334 J/g

Plugging in the values gives the value for q:

q = (200 g)x(334 J/g)

q = 66800 J

The heat required to convert 0 °C ice to 0 °C water = 66800 J

Step 3:

Find the heat required to raise the temperature of 0 °C water to 100 °C water.

q = mcΔT

q = (200 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]

q = (200 g)x(4.18 J/g·°C)x(100 °C)

q = 83600 J

The heat required to raise the temperature of 0 °C water to 100 °C water = 83600J

Step 4:

Find the 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 = (200 g)x(2257 J/g)

q = 451400 J

The heat required to convert 100 °C water to 100 °C steam = 451400J

Step 5:

Find the heat required to convert 100 °C steam to 105 °C steam

q = mcΔT

q = (200 g)x(2.09 J/g·°C)[(105 °C - 100 °C)]

q = (200 g)x(2.09 J/g·°C)x(5 °C)

q = 2090 J

The heat required to convert 100 °C steam to 105 °C steam = 2090J

Step 6:

Find total heat energy. In this final step, put together all of the answers from the previous calculations to cover the entire temperature range.

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

Heat_Total = 2090J + 66800 J + 83600 J + 451400 J + 2090 J

Heat_Total =605980 J

The heat required to convert 200 grams of -5 °C ice into 105 °C steam is 605980 J or 605.980 kJ.