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Chapter 17: Problem 46

Before going in for his annual physical, a \(70.0-\mathrm{kg}\) man whose body temperature is \(37.0^{\circ} \mathrm{C}\) consumes an entire \(0.355-\mathrm{L}\) can of a soft drink (mostly water) at \(12.0^{\circ} \mathrm{C}\) . (a) What will his body temperature be after equilibrium is attained? Ignore any heating bythe man's metabolism. The specific heat of the man's body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . (b) Is the change in his body temperature great enough to be measured by a medical themometer?

Short Answer

Expert verified
(a) The final temperature is approximately 36.8掳C. (b) Yes, the change is measurable.

Step by step solution

01

Understand the Problem

We need to find the final equilibrium body temperature of the man after he consumes the cold drink. We will treat the man's body and the drink as a closed system where heat exchange occurs.
02

Identify Known Values

The key values given are:- Mass of the man, \( m_1 = 70.0 \text{ kg} \)- Initial body temperature, \( T_{1i} = 37.0^{\circ} \text{C} \)- Volume of the drink, \( V_2 = 0.355 \text{ L} \) (which is approximately \(355 \text{ g}\) or \(0.355 \text{ kg}\) for water)- Initial temperature of the drink, \( T_{2i} = 12.0^{\circ} \text{C} \)- Specific heat of the body, \( c_1 = 3480 \text{ J/kg} \cdot \text{K} \)- Specific heat of water, \( c_2 = 4190 \text{ J/kg} \cdot \text{K} \)We need to find the final temperature \( T_f \) of the combined system at equilibrium.
03

Set Up the Heat Exchange Equation

Assuming no heat loss to the environment, the heat lost by the man's body (hot system) will equal the heat gained by the drink (cold system):\[ m_1 \cdot c_1 \cdot (T_f - T_{1i}) + m_2 \cdot c_2 \cdot (T_f - T_{2i}) = 0 \]
04

Substitute Known Values and Solve for Final Temperature

Substitute the known values into the heat exchange equation:\[ (70.0 \times 3480 \times (T_f - 37.0)) + (0.355 \times 4190 \times (T_f - 12.0)) = 0 \]Simplify the equation:\[ 243600 \times (T_f - 37.0) + 1489.45 \times (T_f - 12.0) = 0 \]Distribute and combine like terms:\[ 243600T_f - 9013200 + 1489.45T_f - 17873.4 = 0 \]Combine terms:\[ 245089.45T_f = 9031073.4 \]Finally, solve for \( T_f \):\[ T_f = \frac{9031073.4}{245089.45} \approx 36.8^{\circ} \text{C} \]
05

Determine if the Change is Measurable

The initial body temperature was \( 37.0^{\circ} \text{C} \) and the final equilibrium temperature is approximately \( 36.8^{\circ} \text{C} \). This represents a change of \( 0.2^{\circ} \text{C} \). Medical thermometers typically measure changes of \( 0.1^{\circ} \text{C} \) or greater, so yes, this change is measurable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
In the realm of thermodynamics, heat transfer is a fundamental concept that refers to the transfer of thermal energy between different bodies or systems due to a temperature difference. In our example, when the man drinks the colder soft drink, heat flows from his warmer body to the colder drink until both reach the same temperature. This is because heat always transfers from the hotter object to the cooler one to achieve thermal equilibrium.

Heat transfer can occur in several ways, but in this scenario, it happens primarily through:
  • Conduction: Direct transfer of heat between the man's body and the drink across their boundary.
The rate at which heat is transferred depends on the temperature difference and the properties of the substances involved. Understanding this process is key to predicting how and when two systems will reach the same temperature.
Specific Heat
Specific heat is a property that describes how much heat energy is needed to change the temperature of a substance by a certain amount. Specifically, it is the heat required to raise 1 kilogram of a substance by 1 degree Celsius (or 1 Kelvin).

In the problem, we focus on two specific heat capacities:
  • The specific heat of the man's body: 3480 J/kg路K
  • The specific heat of water (the drink): 4190 J/kg路K
These values indicate that water requires more energy than the human body to change its temperature by the same amount. This is crucial when calculating how the heat exchange will balance out. The specific heat tells us how responsive a material is to heating; materials with high specific heats can absorb a lot of heat without much change in temperature.
Thermal Equilibrium
Thermal equilibrium occurs when two interacting systems reach the same temperature and the exchange of heat stops. At this point, there is no net heat flow between them. In our example, the man's body and the soft drink reach thermal equilibrium at a temperature lower than the initial body temperature but higher than the initial drink temperature.

After reaching equilibrium, the final temperature (calculated to be approximately 36.8掳C) no longer changes without external influences. Understanding thermal equilibrium is crucial for predicting the final temperatures in systems where different temperature bodies interact. It helps in numerous practical applications, such as refrigeration, heating, and designing materials to maintain certain temperature conditions.
Temperature Change Measurement
Measuring the change in temperature is an essential aspect of understanding thermal dynamics and equilibrium processes. In this problem, we calculate that the body's temperature decreases from 37.0掳C to approximately 36.8掳C after consuming the drink.

To determine whether such a change is noteworthy, we consider the precision of common medical devices:
  • Medical thermometers are sensitive to changes as small as 0.1掳C.
So, the change of 0.2掳C is detectably significant. This highlights the importance of accurate measurement tools in medical and scientific settings, allowing for precise monitoring and response to temperature fluctuations. Temperature change measurement provides insight into the effectiveness of heat transfer and thermal balance in systems.

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Most popular questions from this chapter

An electric kitchen range has a total wall area of 1.40 \(\mathrm{m}^{2}\) and is insulated with a layer of fiberglass 4.00 \(\mathrm{cm}\) thick. The inside surface of the fiberglass has a temperature of \(175^{\circ} \mathrm{C}\) , and its outside surface is at \(35.0^{\circ} \mathrm{C}\) . The fiberglass has a thermal conductivity of 0.040 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) (a) What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 \(\mathrm{m}^{2} ?(\mathrm{b})\) What electric- power input to the heating element is required to maintain this temperature?

(a) On January \(22,1943,\) the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{F}\) on January \(23,1916 .\) The next day the temperature plummeted to \(-56^{\circ} \mathrm{C}\) . What was the temperature change in Celsius degrees?

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a \(70.0-\mathrm{kg}\) man to cool his body 1.00 \(\mathrm{C}^{\circ}\) ? The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . The specific heat of a typical human body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise \(17.37 ) .\) (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{cm}^{3}\right) .\)

While running, a \(70-\mathrm{kg}\) student generates thermal energy at a rate of 1200 \(\mathrm{W}\) . To maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to \(44^{\circ} \mathrm{C}\) or higher. The specific heat of a typical human body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , shightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

A Thermos for Liquid Helium. A physicist uses a cylindrical metal can 0.250 \(\mathrm{m}\) high and 0.090 \(\mathrm{m}\) in diameter to store liquid helium at 4.22 \(\mathrm{K}\) ; at that temperature the heat of vaporization of helium is \(2.09 \times 10^{4} \mathrm{J} / \mathrm{kg}\) . Completely surrounding the metal can are walls maintained at the temperature of liquid nitrogen, 77.3 \(\mathrm{K}\) , with vacuum between the can and the surrounding walls. How much helium is lost per hour? The emissivity of the metal can is 0.200 . The only heat transfer between the metal can and the surrounding walls is by radiation.
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