phase change problem. from ice, melting, water, boiling, steam?

Question

Calculate the amount of heat required to change 50.0 g ice at -20.0 °C to steam at 135 °C. (Heat of fusion = 333 J/g; heat of vaporization = 2260 J/g; specific heat capacities: ice = 2.09 J/g_K, liquid water = 4.184 J/g_K, steam = 1.84 J/g_K)
(answer: 156 KJ)

But my question is how do you solve for Boiling. My teacher said get the water converted to steam we use heat of vaporization. So what do I do? for water I got 78KJ(0 to 100 degress celsius)

Answer 1

Break the problem down into stages.

The first stage of the heating process described in the problem is heating the ice up from -20° C to its melting point (0° C).
Here we use the formula,
Q = mc(delta T)
Where Q is the heat energy absorbed by the ice, m is the mass of the ice, c is the specific heat of ice, and (delta T) is the change in the ice's temperature.

The next stage in the process is melting the ice. While the ice/water melts, its temperature is constant (at 0° C).
In this stage we use the formula,
Q = m*L
Where, again, Q is the heat energy absorbed and m is the mass of the water, and L is the latent heat of fusion of the ice/water.

The next stage involves heating the recently melting water from 0° C to its boiling point (100° C). Again, we use the formula,
Q = mc(delta T)
Except this time, c is the melting point of the water (not ice).

Next we boil the water into steam as its temperature stays constant) using the formula,
Q = m * L
Where L is not the latent heat of vaporization of water.

And finally, we must heat the steam up from 100° C to its final temperature of 135° C using, again, the formula,
Q = mc(delta T)
Where c is now the specific heat of steam.

To get the total heat required to accomplish this whole process, add the heat energy required at each stage in the process. In the end it comes down to plugging a bunch of numbers into two formulas and adding it all together.


After working through the whole problem, I calculate the answer to be: 155880 Joules of heat, or about 156 kJ as you said it should be.

Answer 2

Break the problem down into stages.

The first stage of the heating process described in the problem is heating the ice up from -20° C to its melting point (0° C).
Here we use the formula,
Q = mc(delta T)
Where Q is the heat energy absorbed by the ice, m is the mass of the ice, c is the specific heat of ice, and (delta T) is the change in the ice's temperature.

The next stage in the process is melting the ice. While the ice/water melts, its temperature is constant (at 0° C).
In this stage we use the formula,
Q = m*L
Where, again, Q is the heat energy absorbed and m is the mass of the water, and L is the latent heat of fusion of the ice/water.

The next stage involves heating the recently melting water from 0° C to its boiling point (100° C). Again, we use the formula,
Q = mc(delta T)
Except this time, c is the melting point of the water (not ice).

Next we boil the water into steam as its temperature stays constant) using the formula,
Q = m * L
Where L is not the latent heat of vaporization of water.

And finally, we must heat the steam up from 100° C to its final temperature of 135° C using, again, the formula,
Q = mc(delta T)
Where c is now the specific heat of steam.

To get the total heat required to accomplish this whole process, add the heat energy required at each stage in the process. In the end it comes down to plugging a bunch of numbers into two formulas and adding it all together.


After working through the whole problem, I calculate the answer to be: 155880 Joules of heat, or about 156 kJ as you said it should be.

Answer 3

You have the water at 100C. It needs more energy to get to steam.
50g water requires 50 times heat of vaporization 2260 J/g. Please remember that the steam now needs to be heated to 135C. (35 times 1.84)
Make assumption of no heat loss and no mass loss.

Answer 4

1st use q=mcp(T2-T1) to go from -20 to 0
q=50g(2.09J/g_K)(20K)
add to this q=heat of fusion*mass
q1=50g(333 J/g)
then use the first equation again to go from 0 to 100
q2=50g*(4.184J/g_K)*(100K)
then add q=heat of vaporation*mass
q3=50g*(2260J/g)
then use the first equation again to go from 100 to 135 degrees
q4=50g*(1.84 J/g_K)*(35K)
Add q1+q2+q3+q4 to get the overall amount of heat required.