Calorimetry ProblemI was doing a problem in thermodynamics where the net heat is 0.I don't understand why if you have say a copper calorimeter with water at say 15 °C and add a mass of copper at a higher temperature say 90 °C that when calculating the final temperature you would use for the copper piece this in the formula: Q = m c ( T f − T i ) Q = m c ( T f − T i )Where f is for final and i is for initial. Say mass is 0.3 kg. I was told that regarding T f − T i I would need to use 90 − T. This problem was to work out the final temperature of the system. Why was the initial 90 °C of the added copper substituted with T f in the heat equation?

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Calorimetry ProblemI was doing a problem in thermodynamics where the net heat is 0.I don&#039;t understand why if you have say a copper calorimeter with water at say 15 °C and add a mass of copper at a higher temperature say 90 °C that when calculating the final temperature you would use for the copper piece this in the formula:  Q  =  m  c  (      T          f        −      T          i        )  Q  =  m  c  (      T          f        −      T          i        )Where f is for final and i is for initial. Say mass is 0.3 kg. I was told that regarding       T          f        −      T          i       I would need to use   90  −  T. This problem was to work out the final temperature of the system. Why was the initial 90 °C of the added copper substituted with       T          f       in the heat equation?

Jamal Hamilton · Accepted Answer

Heat lost by copper calorimeter = Heat gained by water  −  m  c  (      T    f    −      T    i    )  =      m    w        c    w    (      T    f    −      T    i    )For thermal equilibrium final temperature for both the components will be the same. Let the equilibrium temperature be T and       15          ∘        C  ≤  T  ≤      90          ∘        C  −  m  c  (  T  −  90  )  =      m    w        c    w    (  T  −  15  )  m  c  (  90  −  T  )  =      m    w        c    w    (  T  −  15  )Its quite easy. When a hot object is kept in contact with a cooler object, the hot thing will lose heat and the cold thing will gain that energy. Therefore, for the hot object   Δ      Q    =  −  v  e and for the cold object   Δ  Q  =  +  v  eTo avoid confusion just use the magnitudes of their respective change in heat.

Flakqfqbq · Accepted Answer

This is the concept based on   heat gained  =  heat lost.we basically have      m    1        c    1    (  δ      T    1    )  =      m    2        c    2    (  δ      T    2    )we are now concerned of either heat lost or heat gained. By   (  90  −  T  ) we mean that the temperature of copper has gone down. And by   (  T  −  90  ) we mean the temperature of the body has gone up. That&#039;s the difference. Hope it was helpful

Calorimetry Problem I was doing a problem in thermodynamics where the net heat is 0. I don't und

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