Question

I am currently working 6.45. I had the equation set up corrctly until i got to the circled portion of the second picture. I dont understand why they multiplied the (10J/C) by (Tfinal - 22.8C) when i already had (Tfinal -22.8C) in the equation right before that.

I hope that makes sense. Thank you!

place 6.45 A 455-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 of 10.0 J/K for the vessel, what is the final temperature (c of copper = 0.387 J/g.K)?

Answer 1

When you add the heated copper to the insulated vessel, the temperature of water rises to T_final and that is why the heat gained by water is-

= mass of water x heat capacity of water x rise in temperature

= 159 x 4.184 x (Tfinal - 22.8)

But, not just the temperature of water rises, but the temperature of the vessel also rises. So, vessel also gains some heat. Heat gained by the vessel is-

= heat capacity of vessel x rise in temperature of vessel

= 10.00 x (Tfinal - 22.8)

So, the total heat gained is

= heat gained by the water + heat gained by the vessel

= 159 x 4.184 x (Tfinal - 22.8) + 10.00 x (Tfinal - 22.8)

That is why you see it twice.

Comment if any problem