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Relationship of Internal Energy, Heat, and Work

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    The air inside of a blimp is heated to create lift. The blimp as a system expands to do 81 kJ of work, and absorbs 150 J of heat,. What is the change of the internal energy of the system? (Answer to be provided in kJ)

    ANSWER: -80.85 (kJ)

    STEP ONE: Identify the variables that we are given, which are: Heat (q) and Work (w). Where q=150 (J) and w=81 (kJ)

    STEP TWO: Identify what we are searching for, which is: Change in Internal Energy \[\Delta E\].

    STEP THREE: Realize that the equation required for this problem is \[\Delta E=q+w\]

    STEP FOUR: The problem requests for an answer in (kJ), so we need to convert Heat (q) from (J) to (kJ). This is done through dimensional analysis \[(\frac{150J}{1})*(\frac{1kJ}{1000J})=.15kJ\] So, we have 0.15 (kJ) of Heat (q).

    STEP FIVE: Now that we have our units correct we must check to see whether or not the Work (w) done is negative or positive. Work (w) is positive when work is done ON the system, however Work (w) is negative when work is done BY the system. The question defines the system as the blimp, which means that the work must be negative since the work is done BY the system as the blimp expands. So Work (w) is -81 kJ.

    STEP SIX: Now that we have correct units and determined the work to be negative we can plug the variables into the equation. \[\Delta E=(.15kJ)+(-81kJ)\]. So, \[\Delta E=-80.85kJ\]. Thus, the change in internal energy is -80.85kJ

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