The ice to heavy steam problem is a timeless warm energy homework-related problem. This will outline the procedures crucial to finish this problem and follow up with a worked instance difficulty.

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The amount of warm needed to raise the temperature of a material is proportional to the mass or amount of the product and the magnitude of the temperature readjust.

The equation most typically associated via the warm essential is

Q = mcΔT

whereQ = Heat energym = massc = certain heatΔT = readjust in temperature = (Tfinal– Tinitial)

A good method to remember this formula is Q = “em cat”.

You might notification if the final temperature is reduced than the initial temperature, the heat will certainly be negative. This suggests as the product cools, power is lost by the product.

This equation just applies if the material never before transforms phase as the temperature transforms. Additional warmth is forced to readjust from a solid to a liquid and also as soon as a liquid is changed into a gas. These 2 warmth worths are known as the heat of fusion (solid ↔ liquid) and also heat of vaporization (liquid ↔ gas). The formulas for these heats are

Q = m · ΔHfandQ = m · ΔHv

whereQ = Heat energym = massΔHf= heat of fusionΔHv= warmth of vaporization

The full warm is the amount of all the individual warm adjust procedures.

Let’s put this in exercise with this ice to heavy steam difficulty.

Ice to Steam Problem

Question: How a lot warmth is forced to convert 200 grams of -25 °C ice right into 150 °C steam?Useful information:Specific heat of ice = 2.06 J/g°C Specific warmth of water = 4.19 J/g°CSpecific warmth of steam = 2.03 J/g°CHeat of fusion of water ΔHf = 334 J/gMelting allude of water = 0 °C Heat of vaporization of water ΔHv = 2257 J/gBoiling allude of water = 100 °C

Solution: Heating cold ice to hot vapor calls for 5 distinctive steps:

Heat -25 °C ice to 0 °C iceMelt 0 °C solid ice right into 0 °C liquid waterHeat 0 °C water to 100 °C waterBoil 100 °C liquid water into 100 °C gaseous steamHeat 100 °C vapor to 150 °C steamStep 1: Heat -25 °C ice to 0 °C ice.

The equation to usage for this step is “em cat”


Q1 = mcΔT

wherem = 200 gramsc = 2.06 J/g°CTinitial = -25 °CTfinal = 0 °C

ΔT = (Tfinal– Tinitial)ΔT = (0 °C – (-25 °C))ΔT = 25 °C

Q1 = mcΔTQ1 = (200 g) · (2.06 J/g°C) · (25 °C)Q1 = 10300 J

Step 2: Melt 0 °C solid ice right into 0 °C liquid water.

The equation to usage is the Heat of Fusion warm equation:

Q2 = m · ΔHfwherem = 200 gramsΔHf = 334 J/gQ2 = m · ΔHfQ2 = 200 · 334 J/g Q2 = 66800 J

Step 3: Heat 0 °C water to 100 °C water.

The equation to usage is “em cat” aacquire.

Q3 = mcΔT

wherem = 200 gramsc = 4.19 J/g°CTinitial = 0 °CTfinal = 100 °C

ΔT = (Tfinal– Tinitial)ΔT = (100 °C – 0 °C)ΔT = 100 °C

Q3 = mcΔTQ3 = (200 g) · (4.19 J/g°C) · (100 °C)Q3 = 83800 J

Tip 4: Boil 100 °C liquid water into 100 °C gaseous heavy steam.

This time, the equation to usage is the Heat of Vaporization warm equation:

Q4 = m · ΔHv

wherem = 200 gramsΔHv = 2257 J/g

Q4 = m · ΔHfQ4 = 200 · 2257 J/g Q4 = 451400 J

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

Once aget, the “em cat” formula is the one to use.

Q5 = mcΔT

wherem = 200 gramsc = 2.03 J/g°CTinitial = 100 °CTfinal = 150 °C

ΔT = (Tfinal– Tinitial)ΔT = (150 °C – 100 °C)ΔT = 50 °C

Q5 = mcΔTQ5 = (200 g) · (2.03 J/g°C) · (50 °C)Q5 = 20300 J

Find the full heat

To uncover the full warm of this procedure, add all the individual parts together.

Qfull = Q1 + Q2 + Q3 + Q4 + Q5Qcomplete = 10300 J + 66800 J + 83800 J + 4514400 J + 20300 JQtotal = 632600 J = 632.6 kJ

Answer: The warm essential to transform 200 grams of -25 °C ice into 150 °C steam is 632600 Joules or 632.6 kiloJoules.

The primary point to remember with this form of trouble is to use the “em cat” for the components wbelow no phase readjust occurs. Use the Heat of Fusion equation when changing from solid to liquid (liquid fsupplies right into a solid). Use the Heat of Vaporization as soon as altering from liquid to gas (liquid vaporizes).

Another allude to save in mind is the warmth energies are negative as soon as cooling. Heating a product suggests adding power to the material. Cooling a material implies the material loses energy. Be sure to watch your indications.

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Heat and Energy Example Problems

If you require more instance troubles choose this one, be sure to check out our various other heat and also energy instance problems.

Specific Heat Example ProblemHeat of Fusion Example ProblemHeat of Vaporization Example ProblemOther Physics Example ProblemsGeneral Physics Worked Example Problems