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Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, \(0.6 \mathrm{kg}\) of blood flows to the body's surface and releases \(2000 \mathrm{J}\) of energy. The blood arriving at the surface has the temperature of the body's interior, \(37.0^{\circ} \mathrm{C} .\) Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.

Short Answer

Expert verified
The temperature of the blood that returns to the interior is approximately \(36.2^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the Given Values

We are given the mass of the blood, the amount of energy released, and the initial temperature of the blood. The mass of the blood is \(0.6 \mathrm{kg}\), the energy released is \(2000 \mathrm{J}\), and the initial temperature is \(37.0^{\circ} \mathrm{C}\).
02

Recall the Specific Heat Equation

The specific heat equation is \(Q = mc\Delta T\), where \(Q\) is the heat energy released or absorbed, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. Since blood is assumed to have the same specific heat capacity as water, \(c = 4.18 \mathrm{J/g}^{\circ} \mathrm{C}\). We will use \(c = 4180 \mathrm{J/kg}^{\circ} \mathrm{C}\) for our calculations with the mass in kilograms.
03

Solve for Temperature Change

Rearrange the specific heat equation to solve for \(\Delta T\): \[\Delta T = \frac{Q}{mc}\]Substitute the values: \[\Delta T = \frac{2000 \mathrm{J}}{0.6 \mathrm{kg} \times 4180 \mathrm{J/kg}^{\circ} \mathrm{C}}\]Perform the calculation to find \(\Delta T\).
04

Calculate New Temperature

Using the calculated \(\Delta T\), find the final temperature \(T_{f}\) of the blood as it leaves the surface:\[T_{f} = T_{i} - \Delta T\]where \(T_{i} = 37.0^{\circ} \mathrm{C}\). Substitute the calculated \(\Delta T\) and solve for \(T_{f}\).
05

Verify Calculation

Calculate: \[\Delta T = \frac{2000}{0.6 \times 4180} \approx 0.797^{\circ} \mathrm{C}\]Then, the final temperature:\[T_{f} = 37.0^{\circ} \mathrm{C} - 0.797^{\circ} \mathrm{C} \approx 36.2^{\circ} \mathrm{C}\]

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Key Concepts

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

Heat Transfer
Heat transfer is a core principle in many scientific disciplines and everyday applications, such as cooking or climate control. In the context of blood temperature regulation, heat transfer is the process by which energy is moved from one place to another due to a temperature difference.

Heat can be transferred in several ways:
  • Conduction - directly through materials or substances.
  • Convection - through the movement of liquids or gases.
  • Radiation - via electromagnetic waves, like heat from the sun.
In our scenario, blood carries heat energy from the body's interior to the surface, where it is then released, or transferred, to the environment. This helps to maintain the body's thermal balance by regulating temperature.

When the blood reaches the body’s surface having released heat, its temperature drops, which is essential to cool down the body during activities like exercising.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and the energy of systems. The key principle related to our exercise is the first law of thermodynamics, also known as the conservation of energy.

This law states that energy cannot be created or destroyed, only transferred or converted from one form to another. This principle is why we use the specific heat capacity formula, \( Q = mc\Delta T \), to calculate how much a substance's temperature changes when it gains or loses energy.

The specific heat capacity tells us how much energy is required to raise the temperature of a given mass by one degree Celsius. Since blood is mainly water, we assume blood's specific heat capacity to be the same as water, \( 4.18 \frac{J}{g^{\circ}C} \) or \( 4180 \frac{J}{kg^{\circ}C} \).

This allows us to quantify the change in blood temperature based on the energy it releases or absorbs when traveling to the body's surface.
Blood Temperature Regulation
Blood temperature regulation is a critical physiological process to maintain body homeostasis. When exercising or during stress, the body generates excess heat. The circulatory system plays a significant role in dispersing this heat through blood flow.

Here's how blood temperature regulation works:
  • Blood absorbs heat from the body's muscles and core.
  • It travels to the surface of the skin, where excess heat is released into the environment.
  • Once cooled, the blood returns to absorb more heat, maintaining a consistent cycle.

This process keeps our internal environment stable, significantly impacting thermal comfort and the efficiency of metabolic functions. Understanding the relationship between heat transfer and blood temperature regulation is vital, especially for athletes or individuals working in extreme conditions. Knowing how the body uses thermodynamics to regulate its internal state provides insights into ensuring optimal performance and health.

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Most popular questions from this chapter

On the moon the surface temperature ranges from \(375 \mathrm{K}\) during the day to \(1.00 \times 10^{2} \mathrm{K}\) at night. What are these temperatures on the (a) Celsius and (b) Fahrenheit scales?

An unknown material has a normal melting/freezing point of \(-25.0^{\circ} \mathrm{C},\) and the liquid phase has a specific heat capacity of \(160 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) One-tenth of a kilogram of the solid at \(-25.0{ }^{\circ} \mathrm{C}\) is put into a \(0.150-\mathrm{kg}\) aluminum calorimeter cup that contains \(0.100 \mathrm{kg}\) of glycerin. The temperature of the cup and the glycerin is initially \(27.0^{\circ} \mathrm{C} .\) All the unknown material melts, and the final temperature at equilibrium is \(20.0^{\circ} \mathrm{C} .\) The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

If a nonhuman civilization were to develop on Saturn's largest moon, Titan, its scientists might well devise a temperature scale based on the properties of methane, which is much more abundant on the surface than water is. Methane freezes at \(-182.6^{\circ} \mathrm{C}\) on Titan, and boils at \(-155.2^{\circ} \mathrm{C}\) Taking the boiling point of methane as \(100.0^{\circ} \mathrm{M}\) (degrees Methane) and its freezing point as \(0^{\circ} \mathrm{M},\) what temperature on the Methane scale corresponds to the absolute zero point of the Kelvin scale?

(a) Objects A and B have the same mass of \(3.0 \mathrm{kg}\). They melt when \(3.0 \times 10^{4} \mathrm{J}\) of heat is added to object \(\mathrm{A}\) and when \(9.0 \times 10^{4} \mathrm{J}\) is added to object B. Determine the latent heat of fusion for the substance from which each object is made. (b) Find the heat required to melt object A when its mass is \(6.0 \mathrm{kg}\).

Suppose you are selling apple cider for two dollars a gallon when the temperature is \(4.0{ }^{\circ} \mathrm{C} .\) The coefficient of volume expansion of the cider is \(280 \times 10^{-6}\left(\mathrm{C}^{\circ}\right)^{-1} .\) How much more money (in pennies) would you make per gallon by refilling the container on a day when the temperature is \(26^{\circ} \mathrm{C} ?\) Ignore the expansion of the container.

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