A585−g piece of copper 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 for the vessel of 10.0 j/°c, what is the final temperature of the system (c of copper = 0.387 j/g·°c)?

The final temperature of the system is 32,2°C

Explanation:

The heat losed by copper tubing is equal to heat gained by water and vessel, thus:

(1)

Qcu = 0,387J/g°C × 285g × (X-89,5°C) = 110,295X - 9871,4

Qvessel = 10,0 J/°C (X-22,8°C) = 10,0X - 228,0

QH₂O = 4,184 J/g°C × 159g × (X-22,8°C) = 665,3X - 15167,8

Where X is the final temperature

Using (1):

-110,295X + 9871,4 = 10,0X - 228,0 + 665,3X - 15167,8

25267,2 = 785,595 X

X = 32,2°C

I hope it helps!

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