[babyrecept107]

I am having an issue with this problem.

A piece of gold with a mass of 45.5 g and a temperature of 80.5 C is dropped into 192 g of water at 15.0 C. What is the final temperature of the water?

I did the Q₁ = Q₂ equation:
M x C x (t_i - t_f) = M x C x (t_i + t_f)
(45.5g)(0.129J/g C)(80.5 C - t_f) = (192g)(4.18J/g C)(15.0 C + t_f)
5.8695(80.5 C - t_f) = 802.56(15.0 C + t_f)
472.495 - 5.8695t_f = 12038.4 + 802.56t_f
808.4295t_f = 11565.905
t_f = 14.31 C

That cannot be the right answer because the answer has to be between 15.0 and 80.5 .
PLEASE HELP because this question has been bothering me for a week.

[jusy1]

Your equation should be:

M x C x (tf - ti) = - M x C x (tf - ti)

On the left side you put heat given away by the piece of gold and on the right side the heat absorbed by water:

M x C x (tf - 80.5) = - M x C x (tf - 15)

[babyrecept107]

It gives me the same answer...

[DrCMS]

The correct equation should be


M_Au x C_Au x (80.5-Tf) = M_H20 x C_H20 x (Tf - 15)

[babyrecept107]

M x C x (80.5 - t_f) = M x C x (t_f - 15)
(45.5)(0.129)(80.5 - t_f) = (192)(4.18)(t_f - 15)
5.8695(80.5 - t_f) = 802.56(t_f - 15)
472.495 - 5.8695t_f = 802.56t_f - 12038.4
12510.895 = 808.4295t_f
15.48 C = t_f

I still don't think that's right. The temperature of the gold wouldn't dramatically drop like that... Would it?

[Dan]

Sounds fine. Remember there is ~4 times as much water (by mass) as gold, and the heat capacity of water is ~30 times that of gold. You would expect the temperature of the gold to drop far more than the temperature of the water would rise.

[babyrecept107]

Thank you very much.