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?

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

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.