Question

A pure gold ring (c = 0.128 J/g°C) and pure silver ring (c = 0.235 J/g°C) have a total mass of 16.891 g . The two rings are heated to 66.887 oC and dropped into a 13.9 mL of water at 21.9 oC. When equilibrium is reached, the temperature of the water is 24.2 oC. What is the mass of gold ring? (Assume a density of 0.998 g/mL for water.)

Answer 1

Explanation:

The total mass will be as follows.

          Total mass = mass of silver ring $(M_{s})$ + mass of gold ring $(M_{G})$

And, heat lost by gold and silver rings = heat gained by water

 $[M_{G} \times 0.128 J/g^{o}C \times (66.887 - 24.2)^{o}C] + [M_{s} \times 0.235 J/g^{o}C \times (66.887 - 24.2)^{o}C]$ = $13.9 ml \times 0.998 g/ml \times 4.18 J/g^{o}C \times (24.2 - 21.9)^{o}C$

           $42.687 (0.128M_{g} + 0.235M_{s})$ = 133.368

              $0.128M_{g} + 0.235M_{s}$ = 3.124

             $0.128M_{g} + 0.235 (16.891 - M_{g})$ = 3.124

        $0.128M_{g} + 3.969 - 0.235M_{g}$ = 3.124                

                            $0.107M_{g} = 0.845$

                           $M_{g}$ = 7.897 g

Thus, we can conclude that the mass of gold ring is 7.897 g.