# A gold ring weighing 20 g at 25 °C is dropped into a beaker containing 15 g of boiling water. The final temperature of the water is 97 °C. Recall that the heat capacity of water is 4.184 J/g °C. What is the specific heat capacity of gold?

##### 1 Answer

#### Answer:

#### Explanation:

The first thing to mention here is that water's specific heat at around

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

However, since this is the value given to you in the problem, you must use it.

So, the idea here is that the heat **gained** by the metal will be equal to the heat **lost** by the water. The final temperature of the water will thus be **equal** to the final temperature of the ring.

Mathematically, this is written as

#color(blue)(-q_"water" = q_"metal")#

Here the negative sign is used because **negative**, since it represents heat *lost* by the water.

The equation that establishes a relationship between heat lost/gained and temperature change looks like this

#color(blue)(q = - m * c * DeltaT)" "# , where

*final temperatue* minus the *initial temperature*

In your case, you would have

#- overbrace(m_"water" * c_"water" * DeltaT_"water")^(color(blue)(=q_"water")) = overbrace(m_"metal" * c_"metal" * DeltaT_"metal")^(color(red)(=q_"metal"))#

Rearrange this equation to solve for

#c_"metal" = -m_"water"/m_"metal" * (DeltaT_"water")/(DeltaT_"metal") * c_"water"#

Plug in your values to get

#c_"metal" = (15color(red)(cancel(color(black)("g"))))/(20color(red)(cancel(color(black)("g")))) * ((97 - 100)color(red)(cancel(color(black)(""^@"C"))))/((97 - 25)color(red)(cancel(color(black)(""^@"C")))) * 4.184"J"/("g" ""^@"C")#

#c_"metal" = 0.13075"J"/("g" ""^@"C")#

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the mass of the ring

#c_"metal" = color(green)(0.13"J"/("g" ""^@"C"))#