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 16.884 g . The two rings are heated to 66.895 oC and dropped into a 13.909 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.)

Enter your answer numerically, in terms of grams and to two significant figures.

Please show how you got the answer thankyou!

Answer 1

Let z be the mass (in grams) of the gold ring

Then 16.884 - z is the mass of the silver ring

Mass = Volume x Density

So, Mass of water = (13.909 mL) x (0.998 g/mL)

                              = 13.88 g H2O

Now,

Heat gained by water = (4.184 J/g·°C) x (13.88 g) x (24.2 - 21.9)°C

                                   = (4.184 J/g·°C) x (13.88 g) x (2.3 °C)

                                   = 133.57 J gained by the water (and lost by the rings)

The specific heats of gold = = 0.128 J/g°C

The specific heats of silver = 0.235 J/g°C

Now,

Heat lost by gold = (0.128 J/g·°C) x (z) x (66.895 – 24.2)°C

                                      = (0.128 J/g·°C) x (z) x (42.695 °C)

                                      = 5.46z J

Heat lost by silver = (0.235 J/g·°C) x (16.884 - z) x (66.895 – 24.2)°C

                                        = (0.235 J/g·°C) x (16.884 - z) x (42.695 °C)

                                      = 10.03 (16.884 - z) J

                                      = 169.35 – 10.03z

Add the two expressions for heat lost by the metals, and set them equal to the heat gained by the water:

(5.46z) + (169.35 – 10.03z) = 133.57

169.35 – 4.57z = 133.57

4.57z = 169.35 – 133.57

4.57z = 35.78

z = 35.78 /4.57

z = 7.8 g

So, the mass (in grams) of the gold ring = z = 7.8 g (two significant figures)