A 5.00 kg block of ice at –25 oC is converted to steam at 125 oC. How much energy (in kilojoules) is expended in this process? The following data are provided:
a. Specific heat of ice = 2.092 J/g oC
b. Latent heat of fusion = 334.72 J/g
c. Specific heat of water = 4.184 J/g oC
d. Latent heat of vaporization = 2299.36 J/g
e. Specific heat of steam = 2.008 J/g oC
This isn't complicated but it's a 4 or 5 step process. Here is how you do any problem like this. There are really only two formulas you need (3 actually but two are almost the same).
q = heat required to move T from any T to any other T IN THE SAME PHASE is
q = mass x specific heat in that phase x (Tfinal-Tinitial)
For example, for ice at -25 to ice at zero (starts at ice and continues in same phase as ice.
When you get to a phase change (solid to liquid as in ice melting at the melting point of ice) or (liquid to vapor as in liquid water to steam at boiling point of H2O) it is
for m.p., q = mass ice x heat fusion
for b.p., q = mass water x heat vaporization.
Then add the q for each phase to get total q.
Post your work if you get stuck.
I am so confused. I do not get it at all.
What's the problem?
You go from -25 C to zero C(the melting point) with equation 1 (single phase, ice to start and ice to end)
You melt the ice at 0C using equation 2 (heat fusion one) (this is a phase change from solid to liquid)
You go from liquid water at zero C to 100 C with equation 1. (single phase water to begin and water to end)
You boil the water at 100 C to steam at 100 using equation 3 (the heat vaporization one). (another phase change from liquid to vapor)
You go from steam at 100 to steam at 125 C using equation 1 (single phase of steam to start and steam to end).
Then add all of the q values for each step together for the total.
For the first part for ice at -25 to ice at zero C you have
(mass ice x specific heat ice x (Tfinal-Tinitial).
You have mass ice in grams, specific heat ice listed in your post, Tf is 0 and Ti is -25. Calculate q1 for that part of the problem and go to step 2; i.e., the melting of ice at zero to liquid water at zero C.
To find the amount of energy expended in the process of converting the block of ice to steam, we need to calculate the energy required for each step and then sum them up.
1. Energy required to heat the ice from -25 oC to 0 oC:
The formula for the energy required to change the temperature of a substance is q = m * C * ΔT, where q is the energy, m is the mass, C is the specific heat, and ΔT is the change in temperature.
So, the energy required to heat the ice is q = 5.00 kg * 2.092 J/g oC * (0 oC - (-25 oC)).
First, we need to convert the mass from kg to g, so 5.00 kg = 5000 g.
Therefore, q = 5000 g * 2.092 J/g oC * (0 oC - (-25 oC)) = 261500 J.
2. Energy required to melt the ice at 0 oC:
The formula for the energy required to melt a substance is q = m * Hf, where q is the energy, m is the mass, and Hf is the latent heat of fusion.
So, the energy required to melt the ice is q = 5000 g * 334.72 J/g = 1,673,600 J.
3. Energy required to heat the water from 0 oC to 100 oC:
Using the same formula as in step 1, the energy required is q = 5000 g * 4.184 J/g oC * (100 oC - 0 oC) = 2,092,000 J.
4. Energy required to vaporize the water at 100 oC:
Again, using the formula q = m * Hv, the energy required is q = 5000 g * 2299.36 J/g = 11,496,800 J.
5. Energy required to heat the steam from 100 oC to 125 oC:
Using the same formula as in step 1, the energy required is q = 5000 g * 2.008 J/g oC * (125 oC - 100 oC) = 500,800 J.
Finally, summing up all the energy values calculated above, we get the total energy expended in the process:
Total energy = q1 + q2 + q3 + q4 + q5
= 261500 J + 1,673,600 J + 2,092,000 J + 11,496,800 J + 500,800 J
= 15,024,700 J
To convert the energy from J to kJ, divide by 1000:
15,024,700 J / 1000 = 15,024.7 kJ
Therefore, the amount of energy expended in this process is 15,024.7 kilojoules.