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Ch 17: Temperature and Heat
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 17, Problem 42a

Before going in for his annual physical, a 70.0 kg man whose body temperature is 37.0°C consumes an entire 0.355-L can of a soft drink (mostly water) at 12.0°C. 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 K.

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1
Identify the principle of conservation of energy, which states that the heat lost by the man's body will be equal to the heat gained by the soft drink until thermal equilibrium is reached.
Use the formula for heat transfer: \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
Calculate the mass of the soft drink. Since the density of water is approximately 1 kg/L, the mass of the 0.355-L soft drink is 0.355 kg.
Set up the equation for heat transfer: \( m_{\text{man}}c_{\text{man}}(T_f - T_{\text{man}}) = -m_{\text{drink}}c_{\text{drink}}(T_f - T_{\text{drink}}) \), where \( T_f \) is the final equilibrium temperature, \( T_{\text{man}} \) is the initial temperature of the man, and \( T_{\text{drink}} \) is the initial temperature of the drink.
Solve the equation for \( T_f \), the final equilibrium temperature, by substituting the known values: \( m_{\text{man}} = 70.0 \) kg, \( c_{\text{man}} = 3480 \) J/kg K, \( T_{\text{man}} = 37.0 \)°C, \( m_{\text{drink}} = 0.355 \) kg, \( c_{\text{drink}} = 4186 \) J/kg K (specific heat of water), and \( T_{\text{drink}} = 12.0 \)°C.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). It is a property that indicates how much energy a material can store. In this problem, the specific heat of the man's body is given as 3480 J/kg K, which is crucial for calculating the heat exchange between the man's body and the soft drink.
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Heat Transfer and Thermal Equilibrium

Heat transfer is the process of energy moving from a warmer object to a cooler one until thermal equilibrium is reached, meaning both objects attain the same temperature. In this scenario, the man's body and the soft drink will exchange heat until they reach a common temperature. Understanding this concept helps in setting up the equation to find the final equilibrium temperature.
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Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of this problem, the heat lost by the man's body will equal the heat gained by the soft drink, assuming no other heat exchanges occur. This principle is used to set up the equation that allows us to solve for the final body temperature.
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