A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature. The specific heat of copper (0.385 J/goC) and water (4.18 J/goC) and density of water (1.00 g/mL) will be needed.

Explanation:

Hello,

In this case, considering that the copper is initially hot and the water is initially cold, we can infer that the heat lost by the copper is gained by the water as follows:

Which can be written in terms of temperatures, masses and heat capacities (390 mL equals 390 g for water).

In such a way, solving for the final temperature we obtain:

Regards.

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