Chapter 17: Problem 42
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
Step by step solution
Identify Initial Conditions
Calculate Mass of the Drink
Setup Heat Exchange Equation
Substitute Known Values
Solve for Final Temperature \(T_f\)
Calculate Change in Body Temperature
Determine if Change is Measurable
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Specific Heat
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.
Heat Exchange
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: \( m_2 c_2 (T_2 - T_f) \).
Equilibrium Temperature
To find this temperature, we can use the heat exchange equation:
- \( 70 \times 3480 \times (T_f - 37.0) = 0.355 \times 4186 \times (12.0 - T_f) \)
Energy Conservation
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.