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Questions and Answers

What is the principle that allows the calculation of the specific heat of an object using a calorimeter?

  • The total energy of the system is conserved. (correct)
  • The calorimeter does not need insulation.
  • The mass of the object is always greater than the mass of the water.
  • The object loses all its heat to the environment.
  • In a calorimeter, what can be assumed if no heat is lost to the surroundings?

  • The mass of the calorimeter must be included in calculations.
  • The calorimeter will explode due to excess heat.
  • The temperature of the lead will always be higher than that of the water.
  • The energy absorbed inside must equal the energy released. (correct)
  • What is the specific heat of the lead used in the given example?

  • 0.395 J/goC
  • 0.130 J/goC (correct)
  • 1.000 J/goC
  • 0.250 J/goC
  • Which measurement reflects the initial state of the water in the calorimeter example?

    <p>23.0 °C</p> Signup and view all the answers

    What is the expected outcome of placing a hot object in water within a calorimeter?

    <p>The object and water temperatures will equalize.</p> Signup and view all the answers

    Study Notes

    Specific Heat

    • A calorimeter is a lightweight insulated flask containing water, used to measure the specific heat of objects
    • How it works: An object is placed in a calorimeter, allowing it to reach thermal equilibrium with the water. By analyzing the temperature changes of both the object and the water, you can calculate the specific heat of the object.
    • Key Principle: Energy is conserved. This allows us to calculate the specific heat of the object in the calorimeter.
    • Coffee-cup Calorimeter: A common type of calorimeter used under constant atmospheric pressure.

    Heat Transfer in Calorimetry

    • Energy Conservation: The energy absorbed inside the calorimeter must be equal to the energy released inside it, i.e., qabsorbed = - qreleased.
    • Equation for Heat Transfer: q = m x C x ΔT where:
      • q is the heat energy transferred
      • m is the mass of the substance
      • C is the specific heat of the substance
      • ΔT is the change in temperature

    Specific Heat Examples

    • Example 1: A 75.0g piece of lead (specific heat = 0.130 J/g°C) initially at 435°C is placed in 125.0g of water at 23.0°C.
      • The final temperature of the mixture will be 30.5°C.
    • Example 2: A 97.0g sample of gold at 785°C is dropped into 323g of water at 15.0°C. Gold has a specific heat of 0.129 J/g°C.
      • The final temperature of the mixture will be 22.1°C.

    Homework Examples

    • HW #2: 59.0g of water at 13.0°C is mixed with 87.0g of water at 72.0°C
      • The final temperature of the system will be 48.2°C.
    • HW #4: 240g of water initially at 20.0°C is mixed with an unknown mass of iron at 500.0°C (CFe = 0.4495 J/g°C). The final temperature is 42.0°C.
      • The mass of the iron is 107g.

    Cold Therapy

    • Benefits:
      • Decreases inflammation: Cold constricts blood vessels, reducing blood flow and inflammation.
      • Slows joint fluid production: Excessive synovial fluid can contribute to swelling and discomfort.
      • Distracts the brain from inflammation: Cold stimulates sensory receptors in the skin, decreasing pain signals to the brain.
    • Application: Cold packs should be applied for no more than 20 minutes at a time, but can be repeated throughout the day.

    Heat Therapy

    • Primary Effects:
      • Increased metabolism leading to blood vessel relaxation (vasodilation).
      • Increased blood supply to cool down the heated area.
    • Methods:
      • Conductive: Direct application of heat through hot towels, heating pads.
      • Radiation: Infrared lamps.
      • Radio wave heating (Diathermy): Produces deep heating directly in tissues.
      • Ultrasonic heating: Uses high-frequency sound waves to heat the tissue.

    Ultrasonic Heating

    • Mechanism: Unlike electromagnetic waves, ultrasonic waves cause mechanical vibration in tissues, increasing kinetic energy and leading to heating.

    Energy Expenditure: Fat Loss

    • Physical Activity:
      • Example: A person wishing to lose 4.54kg of fat through physical activity will need to work at 15kcal/min for 2810 minutes (47 hours), assuming a maximum energy expenditure of 9.3kcal/g of fat.
    • Dieting:
      • Example: To lose 4.54kg of fat by reducing food intake, a person consuming 2500kcal/day will need to diet for 84.4 days.

    Basal Metabolic Rate (BMR)

    • Determination: BMR is determined by measuring oxygen consumption at rest.
    • Use: Can be used to determine the energy expenditure of various organs and estimate food energy used during physical activities.
    • Measurement: Oxygen consumption, carbon dioxide production, and heat production can be measured to assess BMR.

    Body Mass Index (BMI)

    • Definition: BMI is a measure of body fat based on height and weight.
    • Significance: BMI does not diagnose health conditions but can indicate an individual's risk level for certain health problems.
    • Range: BMI values span from underweight to obese.

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