m1c1T1=m2C2T2 Help please?

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m1c1T1=m2C2T2 Help please?

Chemistry
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I have the given information, I need help setting up the problem!
The regular formula is: q = m x C x (Tf - Ti) Where: q = amount of heat energy gained or lost by substance m = mass of sample C = heat capacity (J oC-1 g-1 or J K-1 g-1) T = temperature in Celsius q lost= q gained m x C x (Tf - Ti) = m x C x (Tf - Ti)
\[m_1C_1T_1 = m_2C_2T_2\] solving for \(m_2\)\[m_2 = \frac{m_1C_1T_1}{C_2T_2}\] or, solving for \(C_2\)\[C_2 = \frac{m_1C_1T_1}{m_2T_2}\] or, solving for \(T_2\)\[T_2 = \frac{m_1C_1T_1}{m_2C_2}\]

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m1c1T1=m2C2T2 m1 is your mass of copper c1 is the specific heat capacity for copper (which is what you want to find) T1 will be the change in temperature for the copper (final temperature minus initial temperature) m2 is mass of water c2 is the specific heat for water (4.18) T2 is the change in temperature for the water (final temperature minus initial temperature)
@taramgrant0543664 Thanks that helps a ton! just one last question, how would i rearrange the formula to get C1?
So to get c1 by itself you just divide by m1T1 and if you do it on one side you do it to the other so c1= (m2c2T2) / (m1T1) Sorry I'm on my iPad I don't have that fancy equation stuff on here or it would look nicer but I think it gets the point across let me know though if you don't understand!
okay I got it thanks :)

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