Question

A pure gold ring and pure silver ring have a total mass of 15.4 g . The two rings are heated to 68.0 ∘C and dropped into a 13.0 mL of water at 21.5 ∘C. When equilibrium is reached, the temperature of the water is 23.7 ∘C.

Part A

What is the mass of gold ring? (Assume a density of 0.998 g/mL for water.)

Express your answer to two significant figures and include the appropriate units.

mAu=

Part B

What is the mass of silver ring? (Assume a density of 0.998 g/mL for water.)

Express your answer to two significant figures and include the appropriate units.

mAg=

Answer 1

Solution:

Mass of gold and silver ring = 15.4 g

Initial T of rings = 68.0 ⁰C

Volume of water = 13.0 mL

Initial T of water = 21.5 ⁰C

Final T of water = 23.7 ⁰C

Lets calculate heat required to raise the T of water from 21.5 to 23.7 deg C

q = m C delta T

m is mass in g

C is heat capacity

Delta T = change in T

q = 13.0 mL x (0.998 g/mL) x 4.184 J/g deg C   x (23.7-21.5) deg C

=119.42 J

Lets assume mass of gold ring to be x and that of silver is y

x + y = 15.4 g

x = 15.4 – y

-q(metal rings) = q (water)

q (metal ring) = - 119.42

specific heat of gold = 0.129 J / g deg C

specific heat of silver = 0.233 J/ g deg C

total heat given by rings ( q) = q (gold) + q (silver)

Total heat q = - 119.42 J

-119.42 J = m (gold ) x C X delta T + m (silver ) x C X delta T

-119.42 J = (15.4-y) * 0.129 * (23.7-68.0)

                  +( y x 0.233 *(23.7-68.0)

-119.42 J = [(15.4-y) * ( -5.7147)]+[y *-10.322]

119.42 = (88.0063 – 5.7147 y ) +(10.322y)

31.423 = -5.7147 y + 10.322 y

31.423 = 4.61 y

y = 6.88 g

Mass of silver = 6.88 g

And mass of gold = 15.4 – 6.88 = 8.58 g