A 1.00 -kg iron horseshoe is taken from a forge at 900^∘ C and dropped into 4.00 kg of water at

A 1.00 -kg iron horseshoe is taken from a forge at 900^∘ C...

Key Concepts

Entropy

Entropy is a fundamental thermodynamic property that measures the degree of disorder or randomness in a system. In the context of heat transfer, the change in entropy for a reversible process can be calculated by integrating the heat transfer divided by the temperature (dQ/T) over the process path. This concept is crucial for determining the total change in entropy during energy exchanges between systems.

Second Law of Thermodynamics

The Second Law of Thermodynamics states that in any spontaneous process in an isolated system, the total entropy of the system will increase or remain constant. This law provides the direction in which processes naturally occur and ensures that the overall entropy change (system plus surroundings) is non-negative, guiding the analysis of the entropy change when hot and cold bodies come into thermal contact.

Energy Conservation

Energy conservation is a core principle in thermodynamics which states that the total energy in an isolated system remains constant. In problems involving heat transfer between objects, such as a hot object cooling and a cold object warming, the heat lost by one must equal the heat gained by the other. This requirement allows the determination of the final equilibrium temperature and is essential for setting up the entropy change calculations.

Specific Heat Capacity

Specific heat capacity is the amount of energy required to raise the temperature of one unit mass of a substance by one degree Celsius (or Kelvin). It plays a critical role in calculating the amount of heat exchanged during a temperature change. Different materials, like iron and water, have different specific heat capacities, which influences the calculation of the energy transferred and the subsequent entropy change.

Thermal Equilibrium

Thermal equilibrium refers to the state when two or more objects in thermal contact no longer exchange heat because they have reached the same temperature. Establishing thermal equilibrium is a key step in problems where hot and cold bodies are mixed, as it determines the final temperature at which the energy balance and entropy calculations are made.

Transcript

00:01 So here on this problem we are given that one kg iron horseshoe is taken at a 900 degrees celsius and it is dropped into water at 10 degrees celsius.

00:12 So the initial temperature of the iron piece we have 900 degrees celsius and the mass of iron we have 1 kg and the mass of water we are given as 4kg and the initial temperature of water we are given as 10 degrees celsius now we have to assume that no energy is lost by heat to the surrounding and we have to determine the entropy change of the system that is horseshoe plus the water so we have to calculate delta s of system now first of all it will be calculating a final temperature of the system for which we can apply the conservation of energy that is q lost from the iron that is heat lost from iron would be equals to heat gain by water so heat lost by the iron we have mc delta t where delta t is 900 degrees celsius minus the final temperature of the system is equals to heat gain by water and it would be equals to mc here we have to take for iron mc delta t so delta t we have the final minus initial temperature and initial temperature of water we have 10 degrees celsius so these things you have to take it for water now we can put the values the mass for iron we have one kg so 1 kg multiplied by the specific heat of iron which is 448 jule over kg degrees celsius times 900 minus the final we have to calculate is equals to mass of water we have 4 kg and the specific heat of water we have 4186 jule over kj degrees celsius times t final minus initial we have to calculate and the distinct is 10 degrees celsius that is initial temperature of water so from here we'll get the final temperature tf as 33 .2 degrees celsius if you want the final temperature in kelvin we'll get this value as 306 .3 kelvin now we can calculate delta s of the system now the entropy change of the system that is delta s for system it would be equals to entropy change for iron plus the entropy change for water now entropy change for iron we have mc natural log t final over t initial this is for iron plus mc natural lock t final over t initial and this is for water now you can put the values so mass of iron we have 1 kg and specific heat for iron we have 448 joule over kg degrees celsius times natural log final temperature for iron is 306 .3 kelvin divide by initial temperature for iron which is given us a 900 degrees celsius but we have to take the values in kelvin so this value would be 117 kelvin plus mass for water we have 4k g and the specific heat c for water we have 4186 jule over kg degree kelvin times natural lock t final so t final we have 306 .3 kelvin divide by initial temperature of water which is 10 degrees celsius but we will take the values in kelvin so we have 283 kelvin now we can simplify this and we'll get the entropy change for the system to be 717 jule over kelvin...

Please give Ace some feedback