An athlete whose mass is 70.0 kg drinks 16 oz(453.6 g) of refrigerated water. The water is at a

step 1 In this question, we have an athlete mass 70 KG. The athlete drains 453 .6 grams of refrigerated

An athlete whose mass is 70.0 kg drinks 16 oz(453.6 g) of...

An athlete whose mass is 70.0 $\mathrm{kg}$ drinks 16 $\mathrm{oz}(453.6 \mathrm{g})$ of refrigerated water. The water is at a temperature of $35.0^{\circ} \mathrm{F}$ . (a) Ignoring the temperature change of the body that results from the water intake (so that the body is regarded as a reservoir always at $98.6^{\circ} \mathrm{F} ),$ find the entropy increase of the entire system. (b) What If? Assume that the entire body is cooled by the drink and that the average specific heat of a person is equal to the specific heat of liquid water. Ignoring any other energy transfers by heat and any metabolic energy release, find the athlete's temperature after she drinks the cold water, given an initial body temperature of $98.6^{\circ} \mathrm{F}$ . Under these assumptions, what is the entropy increase of the entire system? Compare this result with the one you obtained in part (a).

Key Concepts

Entropy

Entropy is a measure of the disorder or randomness in a system and quantifies the number of microscopic configurations that correspond to a thermodynamic system’s macroscopic state. In thermal processes, entropy change is calculated as the integral of the added or removed heat divided by temperature, and it provides insight into the directionality and irreversibility of energy transfers.

Specific Heat Capacity

Specific heat capacity is a physical property that indicates the amount of heat required to change the temperature of a unit mass of a substance by one degree. It plays a crucial role in thermal problems because it links the energy exchanged in the form of heat to the temperature change of materials, allowing for the calculation of temperature changes during heat transfers.

Thermal Reservoir

A thermal reservoir is an idealized body with a constant temperature even when it exchanges heat with its surroundings, due to its infinite or very large heat capacity. This concept simplifies calculations in thermodynamics by assuming that any heat exchanged does not appreciably change the temperature of the reservoir, which is often used as a reference in entropy computations.

Energy Conservation in Thermal Processes

Energy conservation in thermal processes refers to the principle that the total energy within an isolated system remains constant, with heat lost by one part of the system being gained by another. This principle is essential in problems involving heat exchange, as it permits the determination of unknown temperatures or other thermodynamic properties by equating the heat lost to the heat gained.

Thermal Equilibrium

Thermal equilibrium is the state reached when two or more systems in thermal contact no longer exchange heat, meaning they have reached the same temperature. This concept is fundamental when analyzing the mixing or interaction of substances at different temperatures, as it guides the calculation of the final uniform temperature reached under energy conservation laws.

Transcript

00:01 In this question, we have an athlete mass 70 kg.

00:07 The athlete drains 453 .6 grams of refrigerated water.

00:14 There are two parts in the question.

00:16 In part a, we need to calculate the entropy increase of the entire system by ignoring the temperature change of the body.

00:27 So there are a few things we need to do.

00:31 Okay, first we need to find, we need to convert 35 degrees fahrenheit into degrees celsius.

00:43 So this is 5 over 9, 35 minus 32 degree c, and you get 1 .67 degree c.

00:55 And then you can convert this to kelvin.

00:58 You get 274 .82 kelvin and then 98 .6 degrees fahrenheit.

01:09 If you do the conversion, you get 37 degrees c and then converts to kelvin, you have 310 .15 kelvin.

01:24 Okay so the total the delta s of the system the delta s entire system is equal to the delta s of water plus the delta s of athlete okay here we are treating the athlete as a reservoir okay right so delta as of water is equal to integral of dq over t and then dq is mc d t then divide by t from t i to t f okay so we have 0 .4536 kg the c specific heat capacity of water is for 1 it's 6 and then natural law of t f which is 310 0 .15 divided by ti which is 274 .82.

02:44 You calculate and you get 230 jews per kelvin.

02:51 And then the delta s of the athlete, this is equal to minus q over t.

03:00 So q here is mc delta t, so 0 .4536 times the specific heat capacity multiply by delta t so is 310 .15 minus 274 .8's 2 then divide by 310 310 .15 then you calculate you get minus 216 juice per kelvin okay and then so we can calculate the change in entropy of the system entire system so this is equal to 2 .30 minus 216 and you get 13 .32 jews per kelvin.

04:08 Okay, i'm leaving it to 2 decimal place because if you round up to 1 decimal place, you are going to get the same answer as part b.

04:19 So i'm going to leave it as such.

04:23 Okay, so 13 .32 juice per kelvin is the change in antroval.

04:26 Of the entire system.

04:28 Okay, this is the answer in part a.

04:30 Then in parts b, we assume that the entire body is cooled by the drink and that's the average specific heat of the person is equal to the specific heat of liquid water.

04:45 Okay.

04:46 And then we need to find the delta s of the entire system.

04:50 Okay, so in part b, the first thing we need to do is to find the equilibrium temperature or find a final temperature.

05:00 Okay.

05:04 Find a final temperature at equilibrium...

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