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1 kg of ice at -10c is converted to 1 kg of steam at 600f. what is the total number of joules needed to do so?

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Questions like these test your ability to sum up the energy inputs to a substance that also undergoes phase changes.  With this question, you have H2O which transitions from a solid phase (ice) to the liquid phase (water) to the gas phase (steam).
 
H2O (s) => H2O (l) => H2O (g)
 
The amount of energy as measured in calories or Joules or other forms of the units of energy (BTUs?) that is required either to raise the temperature of the molecule in its present phase or to cause a phase change is known for water by a series of constants or factors.  You need to get each of these values in order to determine the energy input, which also a function of the amount (or mass) of the water (in grams or moles or whatever).  Thus, you need to know:
 
1. amount of energy needed to raise 1 gram solid H2O (ice) by 1° C
2. amount of energy needed to cause 1 gram of solid H2O (ice) to become liquid H2O (water) at its phase change temperature, which is 0° C 
3. amount of energy needed to raise 1 gram of liquid H2O (water) by 1° C
4. amount of energy needed to cause 1 gram of liquid H2O (water) to become gaseous H2O (steam) at its phase change temperature, which is 100° C
5. amount of energy needed to raise 1 gram of gaseous H2O (steam) by 1° C
 
Thus you need to find five constants known specifically for the properties of water.  For instance, you already know list item 3:  most remember that it takes 1 calorie of energy/heat to raise 1 gram of water by 1° C.  This is the specific heat capacity for water.  It has the value with dimensions of 1 cal / g °C.  List items 1 and 5 will be the same.  For list items 2 and 4, these represent values to cause phase changes:  note that the temperature does not change, since a phase change occurs at a specific temperature point.  For those you will need an enthalpy value...the heat of fusion for the solid=>liquid transition, and the heat of vaporization for the liquid=>gas transition.
 
Before putting in a lot of constants (numbers), let's set up the final equation that produces the answer.  The answer will be the sum of five terms, three representing the specific heat required for each phase, and two being the enthalpy values to cause the two phase changes.
 
For the specific heat inputs, we must be mindful that a temperature change is desired, going from one temperature to the final temperature or to the temperature point that represents a phase change!)
 
Thus,
 
total energy input =
    (mass H2O) * (specific heat of solid H2O) * (T2 - T1) +
    (mass H2O) * (heat of fusion for solid H2O)  +
    (mass H2O) * (specific heat of liquid H2O) + (T3 - T2) +
    (mass H2O) * (heat of vaporization for liquid H2O)  +
    (mass H2O) * (specific heat of gaseous H2O) + (T4 - T3)
 
Note that the mass of H2O is common to all the terms, so it can be made a factor by distributing the terms.  The value T1 = -10°C.  The value T2 = 0°, because you are trying to raise the temperature to a phase change point.  Once the phase change from ice to water is done, we continue raising the temperature of water, and we need to do it to its next phase change point, which is T3 = 100° C.  After converting all water to steam, you continue to raise the temperature, apparently to 600° F.  In chemistry, we will see a lot of the mixed use of dimensions or units (calorie vs. Joules, °C vs. °F, gram vs mole) so we must be prepared to put in conversion factors into our final mathematical expressions.  Finally the mass of water used is
1 kg.  These are the constants I was looking for:
 
1. specific heat of ice = 2.01 J/g·°C
2. heat of fusion of water = 6.00 kJ/mol
3. specific heat of water = 4.18 J/g·°C
4. heat of vaporization of water = 40.657 kJ/mol
5. specific heat of steam = 2.01 J/g·°C
 
 
This is a good opportunity to use conversion factors.  The values of items 1, 3, and 5 give the energy units of Joules (J), which is what we want already.  But the values in kJ/mol should be converted to J/g.  I will add these to the final mathematical expression.
 
So the expression now looks like:
 
energy input =
   (100 kg H2O)(1000 g H2O/1 kg H2O) * {
      (2.01 J / g H2O ·°C) * (0 °C - (-10 °C)) + 
      (6.00 kJ / mol H2O) * (1000 J/kJ) * (1 mol H2O / 18.02 g H2O) + 
      (4.18 J / g H2O ·°C) * (100 °C - 0°C) + 
      (40.657 kJ / mol H2O ) * (1000 J/kJ) * (1 mol H2O / 18.02 g H2O) + 
      (2.01 J/ g H2O ·°C) * ((600°F - 32 F) * (100 °C/180 °F) - 100° C)
   }
 
The Google search box is also a calculator, so when I put just the numbers in the Google search box, with hopefully all the parentheses placed correctly to group the terms, I get this:
 
      (100 * 1000) * (2.01 * 10 + 6.00 * 1000/18.02 + 4.18 * 100 + 40.657 * 1000 / 18.02 +
      2.01 * ((600-32)*(100/180) - 100) ) = 346054535.701 = 3461 e+8
 
Thus the answer is 3.461e8 J in scientific notation.
 
Note that is three significant digits.  Make sure your constants (specific heat values, enthalpies at least have more significant digits than the values related to your problem (the mass of water in your case).  We will assume that temperature values are at least 3 significant digits too.
 
Another very important thing to note here is how all the units/dimensions will "cancel out" when arranged to express a product, and that only values with the same dimensions/units can be summed or subtracted.  Most answers you can get correctly in chemistry by making sure this rule is followed...if you can't arrange it like this, then something is wrong, and make sure that the values have the correct dimensions/units.
 
In the last term of the sum of five terms, note that I have incorporated the Fahrenheit conversion, trying to subtract to get to 0°C, then knowing that 180° on the F scale changes for every 100° on the C scale  (212 - 32)/(100 - 0) for the boiling and melting points of water.   If you remember a formula like C = 5/9(F -32), then 5/9(600-32) = 315.5 and you can just plug that number in.
 
Note that I have incorporated the kJ/mol => J/g conversion for water.  You need to know the formula weight for H2O of course.
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