What is the mass of silver ring? (Assume a density of 0.998 g/mL for water.)

This is towards the "difficult" end of enthalpy problems you'll ever see. I haven't seen a harder enthalpy problem than this on the National Chemistry Olympiad.

Let's find the mass of water in the system for easier calculation,

`12.4mL * (0.998g)/(mL) approx 12.375g`

Moreover, recall,

`q = mC_sDeltaT`

In this case, the heat supplied is coming from the exothermic rings once heated.

And before we begin,

`m_1 + m_2 = 15.9g => m_1 = 15.9g - m_2`

`C_(Au) = (0.1256J)/(g*°C)`, and `C_(Ag) = (0.2386J)/(g*°C)`

Hence,

`-[(15.9g-m_2) * (0.1256J)/(g*°C) * (-44.9°C)+m_2 * (0.2386J)/(g*°C) * (-44.9°C)] = 12.375g * (4.18J)/(g*°C) * 2.4°C`
`therefore m_2 = m_(Ag) approx 6.82g`

Solving that monster will give you the approximate mass of the silver ring.

I hope that helps!

We have two conditions to satisfy

(a) Conservation of mass

`"Mass of Au + Mass of gold = Total mass"`

Let `x = "mass of Au"` and
Let `y = "mass of Ag"`

Then

`bb((1))color(white)(m)x + y = "15.9 g"`

(b) Conservation of energy

`"Heat lost by Au + Heat lost by Ag + Heat gained by water" = 0`

`color(white)(mmmm)q_1 color(white)(mmm)+color(white)(mmmm)q_2color(white)(mmll) + color(white)(mmmmm)q_3color(white)(mmmm) = 0`

`"color(white)(mm)m_1C_1ΔT_1color(white)(ml) +color(white)(mmll)m_2C_2ΔT_2 color(white)(ml)+ color(white)(mm)m_3C_3ΔT_3 color(white)(mm)= 0`

`q_1 = x × "0.1256 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × "(22.8- 67.7)" color(red)(cancel(color(black)("°C"))) = "-5.639"xcolor(white)(l) "J·g"^"-1"`

`q_2 = y × "0.2386 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × "(22.8- 67.7)" color(red)(cancel(color(black)("°C"))) = "-10.713"ycolor(white)(l) "J·g"^"-1"`

`m_3 = 12.4 color(red)(cancel(color(black)("mL"))) × "0.998 g"/(1 color(red)(cancel(color(black)("mL")))) = "12.38 g"`

`q_3 = 12.38 "g" × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1")))"g"^"-1" × "(22.8 - 20.4)" color(red)(cancel(color(black)("°C"))) = "124.3 J"`

`q_1 +q_2 + q_3 = "-5.639"xcolor(red)(cancel(color(black)("J")))·"g"^"-1" - 10.713ycolor(red)(cancel(color(black)("J")))·"g"^"-1" + 124.3 color(red)(cancel(color(black)("J"))) = 0`

`q_1 +q_2 + q_3 = "-5.639"xcolor(white)(l)"g"^"-1" - 10.713ycolor(white)(l)"g"^"-1" + 124.3 = 0`

`"-5.639"xcolor(white)(l)"g"^"-1" - 10.713ycolor(white)(l)"g"^"-1" + 124.3 = 0`

`bb((2))color(white)(m)5.639xcolor(white)(l)"g"^"-1" + 10.713ycolor(white)(l)"g"^"-1" = 124.3`

From 1,

`bb((3))color(white)(m)x = "15.9 g -"y`

Substitute Equation 3 into Equation 2.

`5.639("15.9 g -"y)color(white)(l)"g"^"-1" + 10.713ycolor(white)(l)"g"^"-1" = 124.3`

`89.67 -5.639ycolor(white)(l)"g"^"-1" + 10.713ycolor(white)(l)"g"^"-1" = 124.3`

`5.074ycolor(white)(l)"g"^"-1" = 34.6`

`y = 34.6/"5.074 g"^"-1" = "6.82 g"`

The mass of the silver ring is 6.8 g (to two significant figures).

Here's another approach to do it. I got a final expression of:

`m_(Ag) = -[m_wc_w(DeltaT_w//DeltaT_m) + 15.9c_(Au)]/[c_(Ag) - c_(Au)]`

By conservation of mass,

`m_(Ag) + m_(Au) = 15.9`

By conservation of energy,

`q_(Ag) + q_(Au) + q_w = 0`

As a result,

`m_(Ag)c_(Ag)DeltaT_(Ag) + m_(Au)c_(Au)DeltaT_(Au) + m_wc_wDeltaT_w = 0`

The gold and silver change temperature from the same `T_i` to the same `T_f`, so `DeltaT_(Ag) = DeltaT_(Au) -= DeltaT_m`.

`m_(Ag)c_(Ag)DeltaT_m + (15.9 - m_(Ag))c_(Au)DeltaT_m + m_wc_wDeltaT_w = 0`

Distributing and refactoring, we obtain:

`15.9c_(Au)DeltaT_m + m_(Ag)(c_(Ag) - c_(Au))DeltaT_m + m_wc_wDeltaT_w = 0`

Regroup and solve for the mass of silver:

`m_(Ag) = -[m_wc_wDeltaT_w + 15.9c_(Au)DeltaT_m]/[(c_(Ag) - c_(Au))DeltaT_m]`

One more simplification gives:

`m_(Ag) = -[m_wc_w(DeltaT_w//DeltaT_m) + 15.9c_(Au)]/[c_(Ag) - c_(Au)]`

The mass of water would be:

`"12.4 mL" xx "0.998 g"/"mL" = "12.38 g"`

Plug the known values in to get:

`color(blue)(m_(Ag)) = -("12.38 g"cdot"4.184 J/g"^@ "C" cdot (22.8^@ "C" - 20.4^@ "C")/(22.8^@ "C" - 67.7^@ "C") + "15.9 g" cdot "0.126 J/g"^@ "C" )/("0.239 J/g"^@ "C" - "0.126 J/g"^@ "C")`

`=` `"6.77 g Ag"` `~~` `color(blue)("6.8 g")`

If you use `c_(Au) = "0.1256 J/g"^@ "C"` and `c_(Ag) = "0.2386 J/g"^@ "C"`, you will instead get `"6.83 g"`.