Final Temperature of Copper and Water in Insulated Vessel

[John Ker]

Homework Statement

A 529 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 · K]?

Homework Equations

q = q1 + q2 + q3 + ... = 0

Where q = (Tf - Ti) m c

The Attempt at a Solution

This is the part where I am stuck, knowing how to organize the equation.

Ive gotten this so far,

529 * .387 (Tf - 89.5) = 159 * 4.18 * (Tf - 22.8) + 10(Tf - Ti)

Im not sure how the vessel comes into this.

Thanks!

 

[Borek]

Assume initial temperature of the vessel to be that of water it contained.

 

[John Ker]

Assume initial temperature of the vessel to be that of water it contained.

SO the fomula would become:
529 * .387 (Tf - 89.5) = 159 * 4.18 * (Tf - 22.8) + 10(Tf - 22.8)

Where I then distribute everything out and solve for Tf?
Are the signs correct on both sides, I recall seeing that one side needs to be negative, but I am not sure that is relavent here.

 

[Borek]

No, signs are not OK. As your ΔT is defined as final-initial all heats (lost and gain) should be on one side of the equation, just like you wrote in your opening post.

It is also possible to solve such problems using a different convention, one in which we combine things that get colder (lose heat) on one side of the equation and things the gain heat (get hotter) on another side, then all changes in the temperature are assumed to be positive. Mathematically it is equivalent, can be sometimes easier conceptually. The only thing that really matters is that you stick to one convention and not mix them both.

In this particular case the error would be easy to spot - solve the equation you have listed for Tf, does the result make sense?