Aleš Lombergar

You have to know the heat capacity of gold. I just looked it up and the heat capacity of gold is .1291 J/g-K So for a 20 kg bar we have 20000gm of gold * .1291 = 2582 J/K We need to find the same information for glass .8Kg and the water 2kg so again looking it up on the internet I get

Water (liquid) 4.1813 J/g-K and Glass .84 J/g-K

multiplying the mass of each item by its heat capacity I get

Water 8362 J/K and Glass 672 J/K

Now we know that the Gold is 25C, the Glass is 15C and Water is 25C so we need to solve the following equation to find the equilibrium temperature which MUST be between 25 and 15C ( so 26C is not a possible answer)

2582( 25-T) + 8362(25-T) + 672(15-T) = 0

Solving for T, I get an equilibrium temperature of 24.42C

Since gold’s specific heat (0.1291) is much less than water’s (4.1813), we can expect that the final temperature is closer to 15.0 C than to 25.0 C.
20(0.1291)(25-T) = 2.0(4.1813)(T-15)
Solve for T,
T = 17.4 C
———–
The book’s answer of 26.0 C must be wrong. Or you picked a different problem.

No way it’s 26.0C. The book is wrong.

The final answer will be around 24.8C – 24.9C. I would be willing to put down $10 that says it’s in that range.

Edit:
Qgained by glass = Qlost by water + Qlost by gold

Q = mc(DeltaT)

800g(0.84 J/(g.C)(Teq – 15C) = 2,000g(4.184J/g.C)(25C – Teq) + 20,000g(0.129J/g.C)(25C – Teq)

672J/C(Teq) – 10,080J = 209,200J – 8368J/C(Teq) + 64,500J – 2580(Teq)

11,620J/C(Teq) = 283,780J

Teq = (283,780J / 11,620J)/C

>>>((( Teq = 24.42C )))<<<

I lost $10.

Live in the real world. Who puts 20 kg of gold into a container and then worries about the temperature?????!!!!!