An insulated Thermos contains 130 g of water at 80.0^∘ C. You put in a 12.0 g ice cube at 0^∘

An insulated Thermos contains 130 g of water at 80.0^∘ C....

Key Concepts

Second Law of Thermodynamics

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time; during any irreversible process, the total entropy increases. This key concept ensures that in a system mixing items at different temperatures or undergoing phase changes, the net entropy change must be positive, reflecting the directionality of natural processes.

Thermal Equilibrium and Energy Conservation

This concept involves the idea that when two bodies at different temperatures are brought into thermal contact, heat flows from the warmer to the cooler body until both reach the same temperature. The principle of conservation of energy requires that the heat lost by the hot substance must equal the heat gained by the cold substance, allowing one to calculate the final equilibrium temperature of the combined system.

Specific Heat Capacity

Specific heat capacity is a property of a substance, defining the amount of heat required to raise the temperature of a unit mass by one degree Celsius (or one Kelvin). It is crucial when calculating the thermal energy exchanged as substances experience temperature changes, especially in problems involving water or other fluids.

Latent Heat of Fusion

Latent heat of fusion refers to the amount of heat required to change a substance from a solid phase to a liquid phase at its melting point, without a change in temperature. This concept is key in scenarios where a substance like ice is melting, as it must absorb a fixed amount of energy to undergo the phase transition.

Entropy Change and Integration for Variable Temperature Processes

Entropy is a measure of disorder or randomness in a system, and its change during a process can be calculated using the formula ?S = ?(dQ_rev/T). For processes at constant temperature, the change in entropy is simply the heat exchanged divided by the temperature, but for processes involving changing temperature, integration is needed. This concept is critical when calculating the entropy changes associated with both phase changes and temperature variations in a system.

Transcript

00:01 Now in this question we are given this insulated system the system is insulated and it contains water and the mass of water is given to be 130 gram and temperature for water is 80 degrees celsius now we are adding ice which is at zero degrees celsius and the mass is 12 gram into the system like this and then we have to calculate five things after adding this ice to the system now in the a part we have to calculate the equilibrium temperature of the system so we're going to solve the a part first so we know that the system is insulated so there would no no there would be no heat transfer no work transfer so we can write that sigma q will be equals to zero that means we can write that heat gain by water sorry heat lost by water will be equals to heat gain by this ice because ice is at lower temperatures it will go to gain energy heat energy and it will the water will go to lose heat energy so i'm taking that mass of water is 130 gram mass of ice that we are given as 12 gram and the initial temperature for ice is 0 degrees celsius and the initial temperature for water we have 80 degrees celsius this is the input data now we can consider this equation that is delta q equals to zero and we are assuming that the final state of the ice becomes water so first this ice would be converted into water at zero degrees celsius and then from zero degrees celsius to the final temperature that we have to calculate final equilibrium temperature so mass of ice times the latent heat of fusion for ice.

02:09 Now this ice is being converted into water but the temperature is at 0 degrees celsius.

02:15 So plus the mass times this is for ice specific heat times the final temperature minus the initial temperature.

02:25 This is for ice plus the mass of water times the specific heat of water times t final minus the initial and this is for water.

02:38 And that will be equals to 0 now we can put all the values here to find the final temperature that we have to calculate the mass of ice we have 12 gram that means we can write in cages it will be 0 .0 1 2 kgues times the latent heat of fusion for ice it is 3 3 3 jule per kjee plus the mass of ice is 0 .0 1 2 times the specific heat is 4190 times final temperature that we have to calculate minus initial temperature is 0 degrees celsius now plus mass of water is 0 .13kg times the specific heat of water is 4190 multiplied with delta t final temperature we have to calculate an initial temperature we have for water we have 80 degrees celsius and that will be equals to zero so here we have only one variable that is tf and by using our calculator we can easily calculate tf and its value comes out to be as nearly 66 .5 degrees celsius and if you wish to write the answer in kelvin you're going to add 273 to it and we get 339 .6 kelvin 5 kelvin sorry approximately we get this answer in kelvin and this is in degrees celsius so this is the answer for the a part now next we're going to solve the b part of the problem now in the b part we have to calculate the entropic change of the water that was originally ice when it melts now when the ice melts the temperature would remain the same that is zero degrees of this because it is just changing the phase so to calculate and dropy change we can use directly q divided by t now q we have lf times m divide by temperature is 0 degrees celsius means 273 kelvin now we can put the values so latent heat of fusion for ice we have 3 3 3 times 10 to the power 3 multiplied with the mass which is 12 gram means 0 .0 point 012 kg divide by 273 and it comes out to be as nearly 14 .6 jule per kelvin so this is the answer for the b part now next we're going to solve the c part where we have to calculate the entropy change for the water that was originally ice cube but this time we have to calculate entropy when it warms to the equilibrium temperature see, let's initially it was at ice at 0 degrees celsius, then it converted to water at 0 degrees celsius and then from this water it converted, it increases its temperature from 60, from 0 degrees celsius to 66 .5 degrees celsius that we calculated earlier in the a part.

06:14 So we have to calculate entropy change during this time frame.

06:17 This is what it is asking in the c part.

06:21 So this time the temperature is changing.

06:23 So we have to use the integration approach so then tropic change in this time in the c case will be in tropic change will be equals to integration mc delta t divide by the temperature and here temperature is varying from zero degrees celsius means 273 kelvin to 66 .5 degrees celsius that means 33 kelvin 339 .5 kelvin now a we can solve this mc is constant so we can take as outside the integration times natural lock t and it is varying from 273 to 339 .5 kelvin now we can put the values here so after putting the values mass and the specific heat for ice will get the answer to be 11 .0 jule per kelvin we just put mass as 12 gram or in kg.

07:34 It will be 0 .012.

07:36 See we have 4190 and putting the limits we get this answer.

07:42 So i hope you go to the c part of the problem...

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