Textbook 4A.9

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705968765
Posts: 35
Joined: Mon Jan 09, 2023 9:04 am

Textbook 4A.9

Postby 705968765 » Wed Feb 08, 2023 10:34 am

Hi,

I was hoping somebody could walk me through how to solve this problem. I’ve looked at the answer key and it generally makes sense from a mathematical perspective, but doesn’t really explain what is going on conceptually. What equation is being used here to get to the solution?

Thanks!

Anita Sarransingh 2A
Posts: 34
Joined: Mon Jan 09, 2023 8:45 am

Re: Textbook 4A.9

Postby Anita Sarransingh 2A » Thu Feb 09, 2023 8:04 am

Hi! For this problem we are using the heat capacity equation, modified to solve for q, which is q = (delta T)(Heat Capacity). Since the heat capacity gives the value of energy needed to raise the temperature by 1 degree per 1 gram, we would have to multiply the equation by the amount of grams to get the total amount of energy. making the equation q = (delta T)(Heat Capacity)(grams). The heat transferred from each substance will eventually stop when they both reach the same temperature, and since we are assuming that all the heat lost from the metal is being transferred to the water, and all the heat gained from the water came from the metal, then q(water) = -q(metal). In addition, we could rewrite delta T as (final T - initial T). Since we know the initial T values, we would solve for final T.


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