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

Area of CMB = 12

Area of ABC = 24

Step-by-step explanation:

IDK how to prove it exactly but using logic I will go through the steps to findign the answer.

if DM is the midpoint of BC, it splits BC in half which also splites the area of BMC into two equal parts.

If the area of BMD = 6 and CMB = BMD * 2 then CMB = 6 * 2 = 12

Area of CMB = 12

If BM is the midpoint of AC, it splits AC in half which also splits the area into two equal parts.

If the area of CMB = 12 and ABC = CMB * 2 then ABC = 12 * 2 = 24

Area of ABC = 24