Ask Question
22 May, 09:49

A thermally isolated system consists of a hot piece of aluminum and a cold piece of copper. The aluminum and the copper are in thermal contact. The specific heat of aluminum is more than double that of copper. Which object experiences the greater amount of heat transfer during the time it takes the system to reach thermal equilibrium?

a) The aluminum experiences the greater magnitude of heat transfer.

b) The copper experiences the greater magnitude of heat transfer.

c) Both experience the same magnitude of heat transfer.

+3
Answers (1)
  1. 22 May, 12:37
    0
    Make up a question.

    The only change is going to be c.

    Suppose they aluminum starts our higher at 50oC

    Suppose the copper starts out at 20oC

    Suppose the mass of both are 25 grams.

    Aluminum

    m*2c * deltat

    deltat = 50 - x c = 2*c m = 25

    Copper

    m*c*deltat

    deltat = x - 20

    m = 25

    c = c

    Now since the amount of heat is the same (this starts out on a heated slab of something).

    m*2c * (50 - x) = m * c * x - 20 The m and the c are the same. Cancel them out.

    2 * (50 - x) = (x - 20) Remove the brackets.

    100 - 2x = x - 20 Add 20 to both sides.

    120 - 2x = x Add 2x to both sides.

    120 = 3x Divide by 3

    x = 40

    What does this tell you?

    It tells you that the temperature of the aluminum is only going to drop 10 degrees

    The copper is going to gain 40 - 20 = 20 degrees.

    The heat transfer is actually the same. It doesn't take as much heat to heat copper as it does aluminum. That's shown by the difference in how the temperature changes. One looses 10 degrees. The other gains 20. The transfer is the same because of the way the "c" operates.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A thermally isolated system consists of a hot piece of aluminum and a cold piece of copper. The aluminum and the copper are in thermal ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers