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fiasKO [112]
3 years ago
6

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. Assumi

ng 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)?
Chemistry
1 answer:
postnew [5]3 years ago
7 0

Answer:

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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Explanation:

Hello!

In this case, since the reaction between chromate anions and hydrogen ions yields dichromate anions, water and heat, we can infer this is an exothermic reaction by which heat is released (remember in endothermic reactions heat is absorbed as a reactant), it means that considering the LeChatelier’s which states that increasing the temperature of an exothermic reaction shifts the equilibrium leftwards since heat is a product, otherwise (decreasing the temperature) the equilibrium will be shifted rightwards.

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