Chapter 8 problem 21 [ENDORSED]
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Chapter 8 problem 21
Problem 21 states: A piece of copper of mass 20.0 g at 100.0°C is placed in a vessel of negligible heat capacity but containing 50.7 g of water at 22.0°C. Calculate the final temperature of the water. Assume that no energy is lost to the surroundings. Could someone help me out with this problem in solving for T final?
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Re: Chapter 8 problem 21 [ENDORSED]
So we know that the water and the copper will both have the same final temperature, because heat will continue to be exchanged between the two until the temperatures are equal. We also know that, during this process, all of the heat that is lost by the copper will go to the water, because we assume that all heat lost by the system goes into the surroundings. So, given the specific heat capacities in Table 8.2 of .38 Joules per degree celsius per gram of copper and 4.184 Joules per degree celsius per gram of liquid water, we can set up the equation that gives us heat lost by copper=-heat gained by water. So multiply (mass of copper)(T final - T initial of copper)(specific heat capacity of copper)=-(mass water)(T final - T initial of water)(Specific heat capacity of water). Then, with T final as your variable, solve the equation!
Sorry I had to type everything out; I'm not sure how to do all of the symbols on here! But hopefully it made sense!
Sorry I had to type everything out; I'm not sure how to do all of the symbols on here! But hopefully it made sense!
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Re: Chapter 8 problem 21
So, I have a small question regarding the signs: how come it's negative on the side of the water and positive on the copper side? I'm just confused because wouldn't the side of the copper be negative since it's doing work on its surroundings?
Everything else makes sense, though. I'm just a bit confused on the signs. Thanks.
Everything else makes sense, though. I'm just a bit confused on the signs. Thanks.
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Re: Chapter 8 problem 21
Because the heat lost by the system is all gained by the surroundings, the numerical value on each side will be the same, they will just have opposite signs because one is losing energy and the other is gaining it. Because of this, it does not matter which side we place the extra negative sign on, it is only used for the purpose of making the signs equal. So, in this case, the copper will be negative because the final temperature is less than the initial temperature, and the water will be positive because the final temperature is greater than the initial temperature. The extra negative sign was added only to make the signs equal on both sides. If it were placed on the copper side, it would still work because it would just change the overall value from negative to positive, which would match the sign of the water and still work. Hopefully that made sense! :)
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