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

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}