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

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

An athlete whose mass is 70.0 kg drinks 16 ounces $(453.6 \mathrm{g})$ 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 the entire body is cooled by the drink and 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? State how this result compares with the one you obtained in part (a).

An athlete whose mass is 70.0 kg drinks 16 ounces $(453.6 \mathrm{g})$ 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 the entire body is cooled by the drink and 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? State how this result compares with the one you obtained in part (a).

Key Concepts

Entropy Change

Entropy Change is a measure of the change in disorder or randomness in a system when energy is transferred or transformed. In thermodynamics, it is calculated as the integral of the infinitesimal heat transfer divided by temperature (?Q/T). This concept is vital for analyzing processes such as heat exchange between objects at different temperatures and for ensuring that the Second Law of Thermodynamics, which dictates that the total entropy of an isolated system can never decrease, is satisfied.

Heat Transfer

Heat Transfer involves the flow of thermal energy from one body or system to another, typically from a region of higher temperature to one of lower temperature. It is a fundamental concept in understanding processes where energy is exchanged, such as when a cold substance is mixed with a warmer one. Calculating the amount of heat transferred is essential in determining temperature changes and resultant entropy changes.

Specific Heat Capacity

Specific Heat Capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree. This property plays a crucial role in relating the energy transfer (heat) to the resultant temperature change in a system. In problems involving mixtures or changes in temperature of materials, knowing the specific heat capacity helps in computing the energy absorbed or released.

Thermodynamic Reservoir

A Thermodynamic Reservoir is an idealized system that can exchange finite amounts of energy without undergoing a significant change in its temperature. In many calculations, one component (such as the human body in some scenarios) is approximated as a reservoir to simplify analysis by assuming that its temperature remains constant during the energy exchange.

Thermal Equilibrium

Thermal Equilibrium refers to the state in which two or more interacting systems have reached the same temperature, resulting in no net heat transfer between them. When substances at different temperatures are mixed, the final uniform temperature is determined from the conservation of energy, with the systems attaining equilibrium after complete thermal mixing.

Irreversibility

Irreversibility is a concept that describes the net increase in entropy during real processes where the energy transfer cannot be completely recovered to restore the system to its original state. It is associated with the inherent inefficiencies in processes like rapid or non-ideal mixing and provides insight into why certain processes generate additional entropy compared to idealized reversible ones.

Transcript

00:01 Hello everyone in this problem it is given an athlete having the mass so mass of the athlete is 70 kg and its body temperature is 98 .6 ferniides so initial temperature of athlete's body is 98 .6 fernides that is to be converted into kelvin so it would be 310 .15 kelvin he is in taking 16 ounce of water that is 453 .6 grams ounce to be converted in gram at initial temperature 355 degrees of fahrenheit so in kelvin it would be 274 .8 to kelvin in the first part if due to in taking the water the temperature of the body of the athlete doesn't change then we have to calculate entropy increase of the entire system increase in entropy of the entire system if there is no change in temperature of body of the athlete let us see here change in entropy of water we will calculate dq by t initial temperature is 274 p point 8 to final becomes temperature of the body of the athlete that is 310 .15.

02:45 Gq is mcdt, mw is mass of water that is 453 .6 gram.

03:02 Specific heat of water is 1 calorie per gram per kelvin.

03:10 When we integrate 1 upon t with respect to t it will be loan of t and substituting limit so 310 .15 divided by 274 .82 so on solving it change in entropy of water would be 54 .86 calorie per kelvin.

03:45 Now change in entropy of athlete that we have to calculate since there is no change in temperature so amount of the heat supplied by the body of the athlete to the water at initial temperature that will remain same it is no change in mw cw delta t upon initial temperature of athlete now mass of water is 453 .6 specific heat is one calorie per gram per kelvin and changing temperature is 310 .15 minus 274 upon 310 .15 it is to be taken negative because this amount of heat is given by the body of athlete.

05:03 On solving it, change in entropy of the athlete would be minus 51 .67 calorie per kelvin so total entropy change of the system change in entropy of cold water change in entropy of athlete 54 .86 minus 51 .67 so change in entropy of system would be 3 .19 calorie per kelvin this is the answer of first part now in second part it is given if temperature of the body will change if temperature of the body will change if temperature of body of athlete change then we have to calculate final temperature heat lost by the athlete is called to heat gain by cold water so from here heat lost by athlete m mass of the athlete specific heat of athlete into changing temperature of athlete.

07:02 Mass of water, specific heat of water and changing temperature of water.

07:09 Mass is 70 kg so it is to be used in gram.

07:14 Specific heat of athlete is given 1 calorie per gram per kelvin.

07:20 Its initial temperature is 310 .15 kelvin minus final temperature of athlete.

07:31 Mass of water is 453 .6 gram.

07:41 Specific heat is one.

07:43 Final temperature minus initial temperature of water that is 274 .8...

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