In a calorimeter can (which behaves thermally as if it were...

In a calorimeter can (which behaves thermally as if it were equivalent to 40 g of water) are 200 g of water and 50 g of ice, all at exactly $0{ }^{\circ} \mathrm{C}$. Into this is poured $30 \mathrm{~g}$ of water at $90^{\circ} \mathrm{C}$. What will be the final condition of the system? Let us start by assuming (perhaps incorrectly) that the final temperature is $T_{f}>0{ }^{\circ} \mathrm{C}$. Then $30 \mathrm{~g})(1.00 \mathrm{cal} /$ $\mathrm{C}\left(T_{T}\right.$ $-6$ Solving gives $T_{f}=-4.1^{\circ} \mathrm{C}$, contrary to our assumption that the final temperature is above $0^{\circ} \mathrm{C}$. Apparently, not all the ice melts. Therefore, $T_{f}=0^{\circ} \mathrm{C}$. To find how much ice melts, we write Heat lost by hot water $=$ Heat gained by melting ice $$ (30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(90^{\circ} \mathrm{C}\right)=(80 \mathrm{cal} / \mathrm{g}) m $$ where $m$ is the mass of ice that melts. Solving this equation yields $m=34 \mathrm{~g}$. The final system has $50 \mathrm{~g}-34 \mathrm{~g}=16 \mathrm{~g}$ of ice not melted.

In a calorimeter can (which behaves thermally as if it were equivalent to 40 g of water) are 200 g of water and 50 g of ice, all at exactly $0{ }^{\circ} \mathrm{C}$. Into this is poured $30 \mathrm{~g}$ of water at $90^{\circ} \mathrm{C}$. What will be the final condition of the system?
Let us start by assuming (perhaps incorrectly) that the final temperature is $T_{f}>0{ }^{\circ} \mathrm{C}$. Then
$30 \mathrm{~g})(1.00 \mathrm{cal} /$
$\mathrm{C}\left(T_{T}\right.$
$-6$
Solving gives $T_{f}=-4.1^{\circ} \mathrm{C}$, contrary to our assumption that the final temperature is above $0^{\circ} \mathrm{C}$. Apparently, not all the ice melts. Therefore, $T_{f}=0^{\circ} \mathrm{C}$.
To find how much ice melts, we write
Heat lost by hot water $=$ Heat gained by melting ice
$$
(30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(90^{\circ} \mathrm{C}\right)=(80 \mathrm{cal} / \mathrm{g}) m
$$
where $m$ is the mass of ice that melts. Solving this equation yields $m=34 \mathrm{~g}$. The final system has $50 \mathrm{~g}-34 \mathrm{~g}=16 \mathrm{~g}$ of ice not melted.

Key Concepts

Latent Heat of Fusion

Latent heat of fusion is the amount of energy required to change a substance from its solid phase to its liquid phase at its melting point, without a change in temperature. It is a crucial concept in phase change problems, where energy is absorbed or released to overcome intermolecular forces rather than to increase temperature.

Thermal Equilibrium

Thermal equilibrium occurs when all parts of a system reach the same temperature, resulting in no net heat flow between the components. This concept is essential in understanding how mixtures and multi-component systems settle into a stable state after energy exchange.

Calorimetry

Calorimetry is the study and measurement of heat transfer in physical and chemical processes. It involves using devices, such as a calorimeter, to measure the energy exchanged as heat between substances or reactions, ensuring that the total energy is conserved.

Specific Heat Capacity

Specific heat capacity is a property of a material that indicates the amount of heat energy required to raise the temperature of a unit mass of the substance by one degree Celsius. It is fundamental in calculating how much energy is needed for a substance to change its temperature during heat exchange processes.

Transcript

0:00 Good day.

00:01 The topic is about heat.

00:03 When a substance absorbs or loses heat, this heat can cause or may cause a change in its temperature, and the heat that is associated with the temperature change of a substance is solved as q equals mc delta t, where q is the heat, m is the mass, c is the specific heat, which value depends on the type of substance, and delta t is the change in the temperature, which can be solved as the difference in the final temperature and the initial temperature.

00:30 Now, it could also be that the heat that is absorbed and lost by the system may not cause the change in its temperature, but rather on its face.

00:39 And that heat is quantified as q equals ml, where l is the latent heat.

00:44 And depending on which type of phase change, the substance undergoes, l can take different names.

00:52 So like, for example, when the substance undergoes vaporization, then l here is the latent heat of vaporization.

00:59 Now let us consider a container that contains 200 grams of water, which here is a, and that is in equilibrium with 50 grams of ice b.

01:11 And in it is placed 30 grams of water c, which is temperature is at 90 degrees celsius.

01:20 Now we wish to find the final condition of the system that is after mixing these hotter water, what will happen to the system that is will there be ice or they're all that with ice all melt or if all melts what would be the final temperature and so on and so forth now since the temperature of since liquid water a is in equilibrium with ice it follows that the temperature of a and b are both equal to zero degrees celsius so let us suppose in this case or rather let us solve for the amount of heat first first, that liquid water that is poured into it, so that's qc, how much heat can it release if its temperature drops from 90 to 0 degrees celsius? so that is if the 30 grams of water that's added cools all the way down to the freezing temperature of water, then how much heat would be released because that heat would be absorbed by the ice and the water that's already in the container.

02:35 So let's solve that.

02:36 Since there's only change in temperature that's involved, then we will use the formula q equals mc delta t or in other words of putting it, c, tf minus ti.

02:52 So this is the heat released by water c.

02:56 So qc is equal to the mass of that water that's put in is 30.

03:00 The specific heat of water let's use the calorie that's one calorie per gram degree celsius and then we assume the temperature drops to 90 from 90 to zero so that's zero minus 90 so the heat that can be lost when the 30 gram sample cools down all the way to zero degrees celsius is negative 2 ,700 cal and let us assume that let us now solve how much ice would melt if this 2 ,700 degrees celsius or 2 ,700 calories is absorbed...

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