An object with a mass of #125 g# is dropped into #750 mL# of water at #0^@C#. If the object cools by #40 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Apr 19, 2016

Answer:

#c_{object}=1.88J/{g\quad^o C}=0.45\text{cal}/{g\quad^o C}#

Explanation:

Because energy is conserved, the heat gained by the water equals the heat lost by the object.

#Q_{object}=-Q_{water}#

The minus sign is there because one object is losing heat while the other is absorbing heat.

The specific heat of water is known, #\quad c_{water}=4.18 J/{g\quad^o C}#

1 mL of water is the same as 1 gram of water so #m_{water}=750 g#

We have that #Q=mc\Delta T#, so

#Q_{object}=-Q_{water}#

#m_{object} \quad c_{object} \quad \Delta T_{object}=-m_{water}\quad c_{water} \quad \Delta T_{water}#

Substitute known values and obtain: #\quad c_{object}=1.88J/{g\quad^o C}#

If you want the units of the answer to be in terms of calories, #\text{cal}/{g\quad^o C}#, use the unit conversion #1 cal=4.18J#

#c_{object}=0.45\text{cal}/{g\quad^o C}#