# chemistry

A hot lump of 46.9 g of copper at an initial temperature of 54.5 Ã‚Â°C is placed in 50.0 mL of H2O initially at 25.0 Ã‚Â°C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water given that the specific heat of copper is 0.385 J/(gÃ‚Â·Ã‚Â°C)? Assume no heat is lost to surroundings.

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1. mc ΔT (copper) = mc ΔT (water)
(46.9 g)(0.385 J/(g.°C))(Tf °C - 54.5 °C) = (50.0 g)(4.18 J/(g.°C))(Tf - 25.0 °C)
Tf = _____ °C

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2. Plumpy. This only works if you treat each side as absolute values.

The better way to work this is to realize the SUM of the heats gained is zero (one gains, one loses).
mc ΔT (copper) + mc ΔT (water) =0

That way, the math works perfectly.

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