77 grams of an unknown metal at 99ᵒC is placed in 225 grams of water which is initially at 22ᵒC. The water is inside a 44 gram aluminum cup with a specific heat of 0.22 cal/gᵒC. The final temperature of the system is 26ᵒC. What is the specific heat of the unknown metal?

The specific heat capacity of the unknown metal is C = 0.6991 J/g°C = 0.1671 cal/g°C

Explanation:

Heat lost by the unknown metal is equal to the heat gained by the water and aluminium cup.

Given,

Mass of unknown metal = 77 g

Initial Temperature of unknown metal = 99°C

Mass of water = 225 g

Initial Temperature of water = 22°C

Mass of Aluminium cup = 44 g

Specific heat capacity of Aluminium cup = 0.22 cal/gᵒC = 0.92048 J/g°C

Final temperature of the setup = 26°C

Note that, specific heat capacity of water = 4.186 J/g°C

Let the specific heat of the unknown metal be C

Heat lost from the unknown metal

= (77)(C)(99 - 26) = (5,621C) J

Heat gained by water

= (225)(4.186)(26 - 22) = 3,767.4 J

Heat gained by Aluminium cup

= (44)(0.92048)(26 - 22) = 162.00448 J

Heat lost by unknown metal = (Heat gained by water) + (Heat gained by Aluminium cup)

5621C = 3,767.4 + 162.00448 = 3,929.40448

5621C = 3,929.40448

C = (3,929.40448 ÷ 5621) = 0.6991 J/g°C

C = 0.6991 J/g°C = 0.1671 cal/g°C

Hope this Helps!!!

Explanation:

Hello,

In this case, for the given equilibrium, the following equality is accomplished for the involved energy transfer in the system:

Thus, in terms of masses, heat capacities and temperatures:

Hence, solving for the heat capacity of the metal:

Best regards.