A 45.5-8 sample of copper at 99.8^∘ C is dropped into a beaker containing 152 g of water at 18.5

A 45.5-8 sample of copper at 99.8^∘ C is dropped into a...

Key Concepts

Conservation of Energy

This principle states that energy cannot be created or destroyed in an isolated system, only transferred from one form to another. In thermal processes, the heat lost by one substance must equal the heat gained by another, making it fundamental to solving problems where several bodies exchange heat.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It allows us to determine how much heat energy is needed for a given temperature change in a substance and is crucial in setting up equations relating energy transfer to temperature change.

Thermal Equilibrium

Thermal equilibrium is the state in which two or more substances in physical contact reach the same temperature, resulting in no net heat flow between them. This concept is essential in calorimetry problems, as it defines the final condition to which all components in the system eventually settle.

Calorimetry

Calorimetry is the study and measurement of heat transfer in physical and chemical processes. By using the specific heat capacities and masses of the substances involved, one can apply the conservation of energy principle to determine unknown temperatures or quantities of heat exchanged when the substances reach thermal equilibrium.

Transcript

00:01 Okay, so in this question, we put copper into water, and they want us to determine the final temperature when the copper and water reaches equilibrium.

00:11 So they tell us the mass of the copper is 45 .5 grams.

00:15 The initial temperature of copper is 99 .8 celsius, and we can look up the specific heat of copper, which is 0 .385 joules per gram times kelvin.

00:25 They tell us the mass of water is 152 grams.

00:28 The initial temperature of water is 18 .5 degrees celsius, and we can look up the specific heat of water is 4 .184 joules per gram times kelvin.

00:38 So the first thing that we need to do is we've got to convert the celsius into kelvin.

00:44 All we're going to do is add 273, and we should get 372 .8 kelvin plus 273, and we get 291 .5 kelvin.

01:02 Now, in this, so we have water, let's say we have water, right? and then we add copper into the solution.

01:14 That means since copper is hotter, copper is going to lose heat and water is going to gain heat, right? but the amount of heat that the copper loses is gained by the water.

01:23 This means that the queue of water, that means the amount of heat gained by the water is equal to the negative of the heat lost by the copper.

01:34 Right so if we were to put them on the same side we would get q water plus q copper is equal to zero right so now uh the equation that we are going to use to solve is q equals mc delta t and we know that the q are and then we can use this equation right so q actually you can use both of these it doesn't really matter but i'm going to use this one instead so we have, give me a second, the cues are equal to, so we have the specific heat of water, 4 .184, joules divided by grams times kelvin, right, times that by the amount of water, which is 152 grams, right? the specific heat grams, oh, i flipped it around, but it doesn't really matter.

02:41 Grams times that by t final minus t initial, which is 291.

02:49 1 .5 kelvin, 291 .5 kelvin.

02:57 And then we add this whole thing to the q of copper...

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