A 455 g piece of bopper tubing is heated to 89.5 C and placed in an insulated vessel containing 159 g of water at 22.8 C. Assuming no loss of water and a heat capacity of 10. J/K for the vessel, what is the final temperature (c of copper = 0.387 J/g*K)?

The final temperature of the setup = 36.6°C

Explanation:

Let the final temperature of the setup be T

Heat lost by the copper tubing = Heat gained by water and the vessel

Heat lost by the copper tubing = mC ΔT = 455 × 0.387 × (89.5 - T) = (15759.61 - 176.1T) J

Heat gained by water = mC ΔT = 159 × 4.186 × (T - 22.8) = (665.6T - 15175.1) J

Heat gained by vessel = c ΔT = 10 × (T - 22.8) = (10T - 228) J

Heat lost by the copper tubing = Heat gained by water and the vessel

(15759.61 - 176.1T) = (665.6T - 15175.1) + (10T - 228)

851.7 T = 31162.71

T = 36.6°C

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