Calculate the heat change at 100^∘ C for each of the following and indicate whether heat was abs

step 1 Problem 45, we are calculating the heat change. So before we do that, let's go ahead and just id

Calculate the heat change at 100^∘ C for each of the...

Calculate the heat change at $100^{\circ} \mathrm{C}$ for each of the following and indicate whether heat was absorbed or released: a. calories to vaporize $10.0 \mathrm{g}$ of water b. joules to vaporize $5.00 \mathrm{g}$ of water c. kilocalories to condense $8.0 \mathrm{kg}$ of steam d. kilojoules to condense 175 g of steam

Calculate the heat change at $100^{\circ} \mathrm{C}$ for each of the following and indicate whether heat was absorbed or released:
a. calories to vaporize $10.0 \mathrm{g}$ of water
b. joules to vaporize $5.00 \mathrm{g}$ of water
c. kilocalories to condense $8.0 \mathrm{kg}$ of steam
d. kilojoules to condense 175 g of steam

Key Concepts

Latent Heat

Latent heat is the amount of energy per unit mass required for a substance to undergo a phase change at constant temperature and pressure. In this context, latent heat of vaporization and condensation represent the energy needed to change water between its liquid and vapor states without changing temperature, highlighting the inherent energy involved in breaking or forming intermolecular bonds.

Heat Change Calculation for Phase Transitions

The calculation of heat change during phase transitions involves multiplying the mass of the substance by the appropriate latent heat value. This fundamental concept allows for determining the energy required to either vaporize or condense a given amount of substance, emphasizing the direct proportionality between the mass undergoing the phase change and the energy involved.

Unit Conversion

Accurate energy calculations depend on proper unit conversions among calories, joules, and their multiples such as kilocalories and kilojoules. This concept ensures that all quantities are expressed in compatible units, facilitating precise computations and comparisons in thermodynamic calculations.

Exothermic vs. Endothermic Processes

In thermodynamics, identifying whether a process is exothermic or endothermic is crucial. Endothermic processes, such as vaporization, absorb heat from the surroundings, while exothermic processes, such as condensation, release heat. Understanding this distinction helps in correctly interpreting the direction of energy flow relative to the system undergoing the phase change.

Transcript

00:01 Problem 45, we are calculating the heat change.

00:04 So before we do that, let's go ahead and just identify if our scenarios are absorbing heat or releasing heat.

00:15 Okay, so for part a and part b, we are vaporizing water.

00:19 So in order to vaporize, we're turning liquid to gas.

00:23 So that means we're putting in heat to the water molecules.

00:26 So putting in heat means that water is absorbing.

00:31 Heat and on the other hand we are having some condensation for part c and part d so we're turning vapor or gas back into liquid okay so what that means for us is your gas is releasing heat so moving on to our calculations we're going to need the value for the heat of vaporization of water.

01:03 So we got 540 calories per gram of water or 2 ,260 joules.

01:17 Okay.

01:18 So depending on which unit we need, um, those are our values.

01:24 Okay.

01:26 So moving on.

01:27 Let's go ahead and start with part a.

01:33 Part a, i got 10 .0 grams of water, right? and i'm vaporizing that, i'm turning into gas.

01:43 And we want it in units of calories, okay? so what i'm going to do here is take this value, 540 calories per gram of water, and multiplying it by 10 .0 grams.

01:57 Okay, so 540 calories per gram of water.

02:04 So as usual, that unit grams cancels, okay? so all that's left for us to do for this part is to pull in our calculator.

02:16 We've got 10 times 540, okay, which of course is 5 ,400.

02:24 So we got 5 ,400 calories.

02:32 That's it for part 8.

02:33 Okay.

02:34 Next, part b.

02:36 Same thing, but this time we have 5 .00 grams, okay, and we want it in joules.

02:43 So that means i'm using 2 ,260 this time.

02:47 So i'm multiplying this by 2 ,260 joules per gram of water.

02:54 Okay, so again, units cancel.

02:59 And let's go ahead and pull out our calculator.