Steam at 100^∘ C is added to ice at 0^∘ C . (a) Find the amount of ice melted and the final temp

Steam at 100^∘ C is added to ice at 0^∘ C . (a) Find the...

Key Concepts

Latent Heat of Fusion

This is the amount of energy required to change a solid, like ice, into a liquid at its melting point without changing its temperature. It is a crucial concept when calculating how much ice can be melted by a given amount of energy, as the energy must overcome the latent heat of fusion to initiate a phase change from solid to liquid.

Latent Heat of Vaporization

This represents the energy needed to convert a liquid into a gas at its boiling point without a change in temperature. In problems where steam condenses, the latent heat of vaporization is released as latent heat of condensation, and this released energy is important for calculations involving phase transitions and thermal energy transfer.

Specific Heat Capacity

This is a property that measures the amount of heat required to raise the temperature of a unit mass of a substance by one degree. In thermal problems involving the warming or cooling of water, this concept is used to quantify how temperature changes occur in liquids when they absorb or release heat.

Energy Conservation in Thermal Processes

The principle of energy conservation states that the total energy in a closed system remains constant. In thermal problems involving phase changes and temperature changes, the energy released by one process (such as condensation or freezing) must equal the energy absorbed by another (such as warming ice or melting it), allowing for the determination of final system conditions.

Transcript

00:01 Today we're looking at what happens when we add steam to a block of ice, specifically steam that's going to start at t sub c of 100 degrees celsius, and ice that's going to start at t sub i of 0 degrees celsius.

00:14 And to start, let's say we have 10 grams of steam, so mass sub steam of 0 .0 kilograms because we always want to change back to sie units to make everything compatible for one.

00:30 We actually calculate and the mass of the ice is going to be 50 grams so 0 .050 kilograms and basically what we want to do here is we want to start with just what we know is that energy is conserved so we can write this basic calorometer equation of q sub cold equals negative q sub hot and all i mean by that is the energy that goes into it is the energy that is lost from the steam that is specifically the energy required to first phase change the ice into liquid water and then raise the temperature of that ice to our final temperature is the same energy that is resulting from the steam condensing into liquid water and then cooling into our final temperature.

01:20 So note there's two processes on both sides.

01:23 Faye changed from ice to water, warm the water to the final temperature, and on the right side, phase changes.

01:31 Change from steam to water and then cooling the water from 100 to our final temperature.

01:38 So first let's take a closer look at the phase changes, which the energy of which is given by, if you recall, plus or minus depending on if it's gaining or losing energy, just the mass times the latent heat of whatever is going on at what temperature.

01:53 So let's start with looking at q some melt, what energy we would need to melt our whole ice block.

01:59 So just mass sub -i, the whole mass of the ice block.

02:04 And this is going to be l -s -of -f, the latent heat of fusion, which for water, you can look up in table 20 .2 is equal to 3 .33 times 10 to the 5th joules per kilogram.

02:23 That's how much energy you would need to change a kilogram of water from ice into the liquid, or how much.

02:32 Or vice versa.

02:36 So we can go ahead and sub in and calculate, and since we know this is just 0 .05 kg, you can go ahead and calculate that this just generates 16 ,650 joules.

02:49 It's how much energy we would need to melt our entire ice block.

02:53 So similarly, let's see how much energy is resultant from our steam turning back into liquid.

02:59 So q subcondense equals the mass of our steam.

03:04 And now it's the latent heat of vaporization, which the same table will tell you equals 2 .26 times 10 to the 6 joules per kilogram.

03:18 And again, we know the mass is just 0 .01 kilograms.

03:23 So you can go ahead and calculate this value to just be 2 .26 times 10 to the 4 joules.

03:31 Joules and let's compare these two energies so we know that this the energy required to melt the block is a lot less than the energy we get out of the steam turning back into the liquid water so we expect the entire ice block to melt with 10 grams of steam now let's look at the energy of the recently melted ice water and the recently condensed water get it heading to that equilibrium final temperature.

04:04 So just our basic q equals m, c is in the specific heat, delta t.

04:12 So q sub i, i'm going to say for the water that was ice, just the mass of the ice, times the specific heat of water, and then the final temperature minus the initial temperature.

04:26 And again, you can, for your textbook or the internet, textbook in table 20.

04:32 To get that specific heat of water, which is that cw equals 4186, and that is in joules per kilogram degree celsius.

04:47 So we have our mass of 0 .050 kg, our specific heat of water, and we know tf minus ti is the initial temperature of the ice block is just zero.

05:03 So it's just tf.

05:05 So it's really just 4186 times 0 .05 times tf, which you can quickly calculate is 209 .3 tf.

05:15 Now normally i don't just like calculate the exact numbers in intermediate calculations.

05:22 I hold the variables to the end.

05:24 But these are really just halves and powers of 10.

05:27 So i'm not losing any sick figs.

05:29 I'm maintaining all my precision.

05:31 I'm not worried about it here.

05:32 So similarly, let's see the energy of the recently condensed water cooling down.

05:42 So that's ms of s, specific heat of water again, mtf minus ti.

05:47 Only this time, ti, this time previously it was zero, this time it's 100.

05:56 So we can go ahead and simplify this into 41 .1 .3.

06:03 86 tf minus 40186 and that's pretty that's also in jewels i guess they're all in jewels so now we have all of our cues and we can go back and throw them together and our original calorometry equation so back to q cold equals negative q hot so we have the q melt plus the qi equals negative q condensed plus qs.

06:50 And we have all these values...

Please give Ace some feedback