If 180 . g of liquid water at 0^∘ C and 18.0 g of steam at 110 . ^∘ C are mixed in an insulated

If 180 . g of liquid water at 0^∘ C and 18.0 g of steam at...

Key Concepts

Conservation of Energy

This principle states that energy cannot be created or destroyed, only transferred or transformed. In thermal mixing problems, the heat lost by one part of the system must equal the heat gained by another, ensuring that the total energy of the isolated system remains constant.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is a fundamental property used to calculate how much heat energy will change the temperature of a substance without a phase change.

Latent Heat of Vaporization (and Condensation)

The latent heat of vaporization is the energy required for a substance to change from a liquid to a gas without a change in temperature, while latent heat of condensation is the energy released when a gas turns back into a liquid. These concepts are crucial in problems involving phase changes, where energy exchange occurs without temperature change.

Phase Change and Thermal Equilibrium

When substances at different temperatures or phases are mixed in an isolated system, they exchange heat until a uniform temperature, known as thermal equilibrium, is reached. Phase change processes, such as condensation and vaporization, play a critical role in this heat transfer, as they involve significant energy exchanges without changes in temperature.

Transcript

00:02 In this question, we're dealing with a certain mass of water at zero degrees celsius.

00:08 It's already liquid water, and we have a certain mass of steam in 110 degrees celsius in an insulated container.

00:16 It's no heat.

00:16 It's lost in the surroundings.

00:18 We want to know what the final temperature will be for this system.

00:22 So, first of all, let's write heat loss is equal to heat gained, because in this case, the heat is being transferred from the hotter.

00:32 Phase to the or the steam to the colder water.

00:39 So then let's determine what exactly is losing the heat.

00:44 So the steam is going to cool to 100 degrees celsius, which is the boiling point of water.

00:52 Then it's going to condense into water.

00:57 So we will have to take that into account.

00:59 Then the heat from the warmer water that has condensed, will be used to heat to cool water, and then this will be equal to the heat gain by the cool water overall.

01:16 So, first of all, the equation for heat loss due to cooling of steam to 100 degrees celsius would just be the q equals mc delta t equation.

01:29 So we're going to write that down.

01:31 Remember, delta t is t -finalized t initial, but actually in this case, because we're doing exchange, of energy between two phases, we actually will want to switch t -final and t -initial.

01:50 So we will do that later.

01:52 And then we're going to write q because m -c -delta -t, or q -equal -h -h vaporization as a heat loss due to condensation of the steam at 100 degrees celsius.

02:11 So we don't have to negate that value.

02:14 We just make sure that we switch the temperatures for the cooling of steam and the other mc -d delta t equation.

02:21 So we can just calculate that.

02:24 And be sure you're using the correct delta h -vave value.

02:29 Then we're going to write the equation for heat, lost by hot water to the warm, to warm the cool water...

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