In a specific heat experiment, 196 g of aluminum at 107^∘ C is mixed with 52.3 g of water at 1

In a specific heat experiment, 196 g of aluminum at 107^∘ C...

Key Concepts

Conservation of Energy and Thermal Equilibrium

This concept involves the principle that the total energy in an isolated system remains constant. In the context of mixing substances at different temperatures, energy is transferred from the hotter substance to the cooler one until they reach a common equilibrium temperature. The process uses the energy balance equation, ensuring that the heat lost by the hot substance equals the heat gained by the cold substance.

Specific Heat Capacity

Specific heat capacity is a material property that indicates the amount of heat required to change the temperature of a unit mass of a substance by one degree. It is crucial in calculating how temperature changes when heat is added or removed. In experiments involving thermal exchange, the specific heat capacities of the participating substances allow calculation of the equilibrium temperature by relating the heat lost and gained.

Entropy and the Second Law of Thermodynamics

Entropy is a measure of the disorder or randomness of a system, and its change during a process can be calculated by integrating the heat transfer over the temperature. In reversible processes, the change in entropy is given by the formula ?S = ?(dQ/T). The second law of thermodynamics states that the net entropy of an isolated system always increases for irreversible processes, and by calculating the entropy change for each component, one can assess the overall change in the system’s entropy.

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that in a specific heat experiment, we have 196 grams of aluminum, 1906 grams, at a temperature that we're going to call ta, and that is equal to 107 celsius degrees.

00:29 And it is meets with water, so the mass of water in this case is 52 .3 grams, and the temperature is equal to 18 .6 celsius degrees.

00:51 Now for part a of this problem, what we need to calculate is the equilibrium temperature that we are going to call simply as t.

01:04 And for that, we use the fat that the sum of the heats for the aluminum and water should be equal to zero at the equilibrium.

01:19 So for that we will have the mass of the aluminum times the specific heat for aluminum times the temperature, of the equilibrium minus the temperature of aluminum, plus the mass of water times the specific heat for water times the temperature at the equilibrium minus the temperature of water, and this is equal to zero.

01:48 Solving for the equilibrium temperature, we will find that that is equal to the mass of the aluminum times the specific heat of the aluminum times the temperature of the aluminum plus the mass of the water times the specific heat of water times the temperature of water over the mass of aluminum times the specific heat for aluminum plus the mass of water times the specific heat for water.

02:29 So what we need to do is to substitute all of these values.

02:33 But first, we need to have everything in kilograms and kelvin.

02:39 So let me put this more here and this to here.

02:46 So we have more space.

02:49 Okay.

02:50 So to change from grams to kilograms, we need to divide that value by 1 ,000.

02:57 So for aluminum, we will obtain 0 .19.

03:04 And for water we will obtain 0 .052 to 3 kilograms of water.

03:13 Now for to transform from celsius degrees to kelvin we need to add to that value 273.

03:22 So from there adding to that value we will obtain 380 kelvin and for the temperature of of water, we will have a value of 291 .6 kelvin.

03:44 So with that said, you can search for the values of the specific heat for aluminum and water, and you can substitute all of these values in here.

03:56 So we have for aluminum, we have a mass of 0 .196 kilograms times the specific heat for aluminum that we know is 900 joules per kilogram per kelvin times the temperature of aluminum, that is 380 kelvin, plus the mass of water, that is 0 .053 kilograms times the specific kit for water, that is 4 ,000 ,000, 190 joules per kelvin and the temperature of water that is 292 kelvin and all of this divided by the mass of aluminum times the specific heat of aluminum plus the mass of water times the specific heat for water 190 joules per kilogram per kelvin.

05:26 So from this, we obtain a temperature of equilibrium of 58 celsius degrees.

05:40 That can be also written in kelvin as 331 kelvin.

05:46 Now, for part b of this problem, we are asked about, the entropy change of for this part b for the aluminum...

Please give Ace some feedback