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.)
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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.)
Answers
Explanation:The total mass will be as follows. Total mass = mass of silver ring [tex](M_{s})[/tex] + mass of gold ring [tex](M_{G})[/tex]And, heat lost by gold and silver rings = heat gained by water [tex][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][/tex] = [tex]13.9 ml \times 0.998 g/ml \times 4.18 J/g^{o}C \times (24.2 - 21.9)^{o}C[/tex] [tex]42.687 (0.128M_{g} + 0.235M_{s})[/tex] = 133.368 [tex]0.128M_{g} + 0.235M_{s}[/tex] = 3.124 [tex]0.128M_{g} + 0.235 (16.891 - M_{g})[/tex] = 3.124 [tex]0.128M_{g} + 3.969 - 0.235M_{g}[/tex] = 3.124 [tex]0.107M_{g} = 0.845[/tex] [tex]M_{g}[/tex] = 7.897 gThus, we can conclude that the mass of gold ring is 7.897 g.
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