Question

Suppose 20.0 g pieces of gold and iron, both initially at 100oC, are added to different containers of water, both initially at 25oC. What combination leads to the highest final temperature?

Answer 1

Answer:

• The final temperature in the container with Gold is 27.49 ⁰C

• The final temperature in the container with Iron is  33.01  ⁰C

Therefore, the highest final temperature is obtained in the container with Iron.

Explanation:

Q = mcΔT

Where;

Q is the quantity of heat gained or lost

m is the mass of the metals or water

c is the specific heat capacity

ΔT is the change in temperature, T₂ - T₁

T₂  is the final temperature and T₁ is the initial temperature

Heat lost by metals at 100°C is equal to heat gained by water at 25°C

$-Q_{metal} = Q_{water}$

$-Q_{Au} = Q_{H_2O} \\-Q_{Fe} = Q_{H_2O}$

Specific heat capacity of water = 4.18 J/g°C

Specific heat capacity of gold = 0.129 J/g°C

Specific heat capacity of iron  = 0.45 J/g°C

⇒For Gold

-20*0.129*(T₂ - 100) = 18*4.18 (T₂ - 25)

-2.58T₂ +258 = 75.24T₂  - 1881

77.82T₂  = 2139

T₂  = 2139/77.82

T₂ = 27.49 ⁰C

⇒For Iron

-20*0.450*(T₂ - 100) = 18*4.18 (T₂  - 25)

-9T₂  +900 = 75.24T₂ - 1881

84.24T₂ = 2781

T₂  = 2781/84.24

T₂ = 33.01  ⁰C

Therefore, the highest final temperature is obtained in the container with Iron.