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

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.