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A 506g piece of molten copper is dropped into 7 litres of water, which is initially at 18oC. What is the resulting temperature of the water and copper once it has reached equilibrium?
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i assumed, that all energy loss by the copper is tranfered to the water, so so use \[Q=mc \Delta T\] to work out copper energy loss. This is what i get: T(melting) = 1084°C c = 0.385 J/g°C Q = 0.506kg x = 385 J/kg°C x (1084°C - 18°C) = 207667.46 J But it doesn't make sense..
Then my friend says that then i have to "reverse to find temperture gain by the water" what does it mean??
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