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

Short Answer

Expert verified
The man's body temperature slightly decreases, likely not enough for a thermometer to measure precisely.

Step by step solution

01

Identify Initial Conditions

We have a 70.0-kg man with an initial body temperature of 37.0°C and he drinks a 0.355-L can of soft drink at 12.0°C. Treat the drink as water with its density and specific heat similar to that of water: density = 1 kg/L and specific heat = 4186 J/kg·K.
02

Calculate Mass of the Drink

Convert the volume of the drink into mass using the density of water: Mass of drink = 0.355 kg (since 1 L of water is approximately 1 kg).
03

Setup Heat Exchange Equation

Using the principle of conservation of energy, the heat lost by the man's body equals the heat gained by the drink:\[ m_1 c_1 (T_f - T_1) = m_2 c_2 (T_2 - T_f) \]where \( m_1 \) is the mass of the man's body, \( c_1 \) is the specific heat of the man's body, \( T_1 \) is the initial body temperature, \( m_2 \) is the mass of the drink, \( c_2 \) is the specific heat of water, and \( T_2 \) is the initial temperature of the drink.
04

Substitute Known Values

Substitute given values into the equation:\[ 70 \, \text{kg} \times 3480 \, \text{J/kg} \cdot \text{K} \times (T_f - 37.0) = 0.355 \, \text{kg} \times 4186 \, \text{J/kg} \cdot \text{K} \times (12.0 - T_f) \].
05

Solve for Final Temperature \(T_f\)

Rearrange and solve the equation for \( T_f \):\[ 70 \times 3480 \times (T_f - 37.0) = 0.355 \times 4186 \times (12.0 - T_f) \]\[ 243600(T_f - 37.0) = 1485.73(12.0 - T_f) \]Solve the above linear equation to find \( T_f \).
06

Calculate Change in Body Temperature

After solving the above equation, calculate \( \Delta T = T_f - 37.0 \). This will indicate how much the body temperature has changed.
07

Determine if Change is Measurable

Evaluate if the change in body temperature is within the sensitivity range of a typical medical thermometer (usually measurable changes are greater than 0.1°C).

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Key Concepts

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

Specific Heat
The specific heat of a substance is a crucial factor in thermodynamics. It is defined as the amount of heat required to raise the temperature of a unit mass of the substance by one degree Celsius (or one Kelvin). In our exercise, the specific heat plays a key role in determining how much energy is needed to change the temperature of the man's body and the soft drink.
For example:
  • The specific heat of the man’s body is 3480 J/kg·K.
  • The specific heat of the soft drink (since it's mostly water) is 4186 J/kg·K.
The specific heat values tell us how efficient each substance is at absorbing heat. A higher specific heat means that the substance can absorb more heat without a large temperature increase. This concept helps us understand the heat exchange process between the man's body and the soft drink.
Heat Exchange
Heat exchange refers to the transfer of thermal energy between substances that come into contact. In this exercise, the man's body and the cold soft drink interact, leading to an exchange of heat as they seek thermal equilibrium.
The principle at work here is that the warmer entity (the man's body) will lose heat, while the cooler entity (the soft drink) will gain it. This transfer continues until both entities reach the same temperature.
The heat exchange can be described mathematically with an equation:
  • The heat lost by the man's body: \( m_1 c_1 (T_f - T_1) \).
  • The heat gained by the drink: \( m_2 c_2 (T_2 - T_f) \).
This equation is derived from the conservation of energy principle, which ensures that the heat lost by the man's body is equal to the heat gained by the drink.
Equilibrium Temperature
Achieving equilibrium temperature is the goal of the heat exchange process. When equilibrium is reached, the temperature between interacting substances balances out. In this scenario, the man's body temperature and the soft drink temperature change until they meet at a common temperature, known as the equilibrium temperature.
To find this temperature, we can use the heat exchange equation:
  • \( 70 \times 3480 \times (T_f - 37.0) = 0.355 \times 4186 \times (12.0 - T_f) \)
By solving this equation, we determine the final temperature \( T_f \) at which the energy gained by the drink equals the energy lost by the man's body. The equilibrium temperature is important because it lets us predict the final state after the heat exchange.
Energy Conservation
Energy conservation is a fundamental principle of physics, especially in thermodynamics. It states that energy cannot be created or destroyed but only transferred or converted from one form to another.
In our case, the principle of energy conservation ensures that the heat lost by the man's body is exactly equal to the heat gained by the soft drink. This principle is crucial for ensuring accurate calculations in exercises like this one.
When applying energy conservation, here is what happens:
  • The man’s body loses a certain amount of thermal energy.
  • The drink gains the same amount of thermal energy.
By ensuring energy conservation, we accurately predict changes in temperature, both scientifically and practically. This helps us understand how heat affects biological systems and everyday interactions, such as drinking a cold beverage.

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Most popular questions from this chapter

BIO While running, a 70-kg student generates thermal energy at a rate of 1200 W. For the runner to maintain a constant body temperature of 37\(^\circ\)C, this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the energy 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\(^\circ\)C or higher. The specific heat of a typical human body is 3480 J / kg \(\cdot\) K, slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to 34.0\(^\circ\)C overnight and rise to 40.0\(^\circ\)C during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400-kg camel would have to drink if it attempted to keep its body temperature at a constant 34.0\(^\circ\)C by evaporation of sweat during the day (12 hours) instead of letting it rise to 40.0\(^\circ\)C. (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 J/kg \(\cdot\) K. The heat of vaporization of water at 34\(^\circ\)C is \(2.42 \times 10{^6} J/kg\).)

What is the rate of energy radiation per unit area of a blackbody at (a) 273 K and (b) 2730 K?

A spherical pot contains 0.75 L of hot coffee (essentially water) at an initial temperature of 95\(^\circ\)C. The pot has an emissivity of 0.60, and the surroundings are at 20.0\(^\circ\)C. Calculate the coffee’s rate of heat loss by radiation.

A Foucault pendulum consists of a brass sphere with a diameter of 35.0 cm suspended from a steel cable 10.5 m long (both measurements made at 20.0\(^\circ\)C). Due to a design oversight, the swinging sphere clears the floor by a distance of only 2.00 mm when the temperature is 20.0\(^\circ\)C. At what temperature will the sphere begin to brush the floor?

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