Question #190223

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

Expert's answer

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/gÂ·Â°C

specific heat of water = 4.18 J/gÂ·Â°C

specific heat of steam = 2.09 J/gÂ·Â°C

**Step 1:**

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

q = mcÎ”T

where

- q = heat energy
- m = mass
- c =Â specific heat
- Î”T = change in temperature

In this problem:

- q = ?
- m = 200 g
- c = (2.09 J/gÂ·Â°C
- Î”T = 0 Â°C - -5 Â°C=5
^{0}C

Plug in the values and solve for q:

q = (200 g)x(2.09 J/gÂ·Â°C)[(5^{0} C]

q = (200 g)x(2.09 J/gÂ·Â°C)x(5 Â°C)

q = 2090 J

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

**Step 2:**

Find the heat required to convert 0 Â°C ice to 0 Â°C water.

Use the formula for heat:

q = mÂ·Î”H_{f}

where

- q = heat energy
- m = mass
- Î”H
_{f}Â =Â heat of fusion

For this problem:

- q = ?
- m = 200 g
- Î”H
_{f}Â = 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 Â°C ice to 0 Â°C water = 66800 J

**Step 3:**

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

q = mcÎ”T

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

q = (200 g)x(4.18 J/gÂ·Â°C)x(100 Â°C)

q = 83600 J

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

**Step 4:**

Find the heat required to convert 100 Â°C water to 100 Â°C steam.

q = mÂ·Î”H_{v}

where

q =Â heat energy

m = mass

Î”H_{v}Â =Â heat of vaporization

q = (200 g)x(2257 J/g)

q = 451400 J

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

**Step 5:**

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

q = mcÎ”T

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

q = (200 g)x(2.09 J/gÂ·Â°C)x(5 Â°C)

q = 2090 J

The heat required to convert 100 Â°C steam to 105 Â°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}Â = Heat_{Step 1}Â + Heat_{Step 2}Â + Heat_{Step 3}Â + Heat_{Step 4}Â + Heat_{Step 5}

Heat_{Total}Â = 2090J + 66800 J + 83600 J + 451400 J + 2090 J

Heat_{Total}Â =605980 J

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

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