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:

#0.13"J"/("g" ""^@"C")#

Explanation:

The first thing to mention here is that water's specific heat at around #100^@"C"# is not actually equal to #4.181 "J"/("g" ""^@"C")#, but with approximately #4.218"J"/("g" ""^@"C")#.

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 #q_"water"# will be 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

#q# - heat lost/gained;
#m# - the mass of the sample;
#c# - the specific heat of the material
#DeltaT# - the change in temperature, defined as the 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"#

#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"))#