A 101 g piece of aluminum (aluminum = 0.900 J/g oC) is heated to 100.1 oC and added to 48.9 g of water at an initial temperature of 16.5oC. what will the final temperature of the mixture be

The final temperature of the mixture is = 42.14 °C

Explanation:

Mass of the aluminium = 101 gm

Specific heat for aluminium = 0.9

Temperature of aluminium = 100.1 °C

Mass of water = 48.9 gm

Specific heat for water = 4.2

Initial temperature of water = 16.5 °C

From energy balance principal when the aluminium piece is heated and dropped in to the water , the final temperature of aluminium & the water becomes equal.

From the energy conservation principal

Heat lost from the aluminium piece = heat gain by the water

⇒   ( - ) =   ( - ) ----------- (1)

Put all the values in equation (1)

⇒ 101 × 0.9 × (100.1 - ) = 48.9 × 4.2 × ( - 16.5 )

⇒ 90.9 × (100.1 - ) = 205.38 × ( - 16.5 )

⇒ 100.1 - = 2.26 ( - 16.5 )

⇒ 100.1 - = 2.26  - 37.28

⇒ 3.26 = 137.38

⇒ = 42.14 °C

Therefore the final temperature of the mixture is = 42.14 °C