A gold block initially at 62.9°C is put into 120 grams of H2O at 25.8°C in a container. The mixture at thermal equilibrium reached a temperature of 28.3°C. What is the mass of gold used?

Answer 1

The heat from the gold block (@62.9º) will flow to the cooler water (@25.8º). We can use the equation

q = mC∆T to solve this problem, but we will need to look up the values for C.

heat gained by water = q = mC∆T

q = heat = ?

m = 120 g

C = specific heat of water = 4.184 J/gº (look it up)

∆T = change in temperature = 28.3º - 25.8º = 2.5º

Solving for q, we have

q = (120 g)(4.184 J/gº)(2.5º) = 1255 J ... This is the heat gained by the water and also the heat lost by gold

heat lost by gold = q = mC∆T

q = 1255 J

m = mass of gold = ?

C = specific heat of gold = 0.1256 J/gº (look it up)

∆T = change in temperature = 62.9º - 28.3º = 34.6º

Solving for m (mass), we have...

m = q / C∆T = 1255 J / (0.1256 J/gº)(34.6º)

m = 289 g

Depending on the value that you find for specific heat of gold, your answer may be different.

Also, since the mass of water (120 g) contains only 2 significant figures, the answer should more correctly be reported as 290 g to 2 sig. figs.

Answer 2

Use q=m⋅c⋅ΔT

find heat transfer with water

set up equation for heat transfer with gold (leaving mass of gold as unknown)

remember that heat loss=heat gain

so q(gold) is same as q(water)

now rearrange equation, and find value for m(gold)

see video attached

Answer 3

Although I can't give you the answer outright, I can suggest a few concepts that you'll need to understand to solve this problem.

1) To solve this problem you need to assume that heat never leaves the system. Heat will, however, "flow" from a warmer object into a colder object.

2) The concept of "specific heat" if critical to solving this.

3) You will need to look up a few constants in order to solve this problem. The question itself doesn't contain this critical information.

I hope this helps provide a solid foundation for answering this question!