course Phy 242 I may have done this wrong but all I saw in this Query Program was the general knowledge question about how to find the unknown specific heat in a closed system. Is this right or did I miss the University Physics problem? If I did forget to do something let me know and I will go back and redo it. ???o??????Q????assignment #002??????????
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00:12:48 query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht Explain how the final temperature of the system is combined with the known initial temperatures and masses of both substances to obtain the unknown specific heat
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RESPONSE --> To do this problem we must first assume that this is a closed system with no energy gain or loss. That means that the 'dQ1 and the 'dQ2 must add up to zero. This means that 'dQ1 + d'Q2 = 0. This means that we can substitute the change in thermal energies with the specific heat, mass, and inital temperatures of each that are known. So, 'dQ1 + 'dQ2 = 0, becomes c1m1'dT1 + c2m2'dT2 = 0 Now if we use this equation and solve for our unknown we get: First, c2m2'dT2 = - c1m1'dT1 Then, c2 = -(c1m1'dT1)/(m2'dT2) To understand this equation one needs to understand that 'dT = (T final for the system - T initial for each substance).
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00:13:08 ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as `dQ = mass * specific heat * `dT. (General College and University Physics students note that specific heat is not in fact strictly constant and for most substances changes with temperature.) We know that what energy one substance loses, the other gains, and we know the specific heat of water , so we can set the two quantities equal and solve for specific heat of the unknown substance. For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently m1 c1 `dT1 = - m2 c2 `dT2. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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RESPONSE --> Understood.
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