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BR gen chem, Section 8, passage XI, question 74

Discussion in 'MCAT Study Question Q&A' started by Student1331, 05.05.12.

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  1. Student1331

    Student1331 2+ Year Member

    Pre-Health (Field Undecided)

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    BR gen chem, Section 8, passage XI, question 74

    Had Metal III been used in Experiment II instead of Metal IV, the final temperatures for the liquids would have been:

    I answered C (the answer is D) because both Metal III and Metal IV are heated up to the same temperature before being placed in the liquid. They are heated to the SAME temperature, which would indicate they have the same energy. In the liquid, the metal with the high heat capacity (metal IV) would give off less heat, while retaining most of it, while the metal with the smaller heat capacity would give off a greater amount of heat, while retaining a smaller amount of heat. Thus the temp for liquid with Metal IV would be lower, and, for liquid with Metal III would be higher.

    What do you guys say?
  2. johnwandering

    johnwandering 5+ Year Member


    You are somewhat confused about the definition of heat capacity. Heat Capacity is the Energy needed to raise the temperature of an object by 1 degree Celsius. Hence, if one object had a very very high heat capacity, it would take a LOT of energy to heat it.

    If one metal has a higher Heat Capacity, it would require considerably MORE energy to heat to temperature X. If both metals are heated to the Same Temperature, the one with the higher heat capacity will have considerably MORE energy than the other.

    So in the end, when you put both metals in a liquid to impart the energy they have to the water, the object with the higher heat capacity will impart a LOT more energy to the liquid.
    (Don't forget, when you place the metal into the liquid, the system WILL EQUALIZE until the metal and the liquid are the same temperature.)
  3. drechie

    drechie 5+ Year Member


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