Question

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

Answer 1

Answer:

The final temperature of the mixture is = 42.14 °C

Explanation:

Mass of the aluminium $m_{a}$ = 101 gm

Specific heat for aluminium $C_{a}$ = 0.9 $\frac{J}{gm c }$

Temperature of aluminium $T_{a}$ = 100.1 °C

Mass of water $m_{w}$ = 48.9 gm

Specific heat for water $C_{w}$ = 4.2 $\frac{J}{gm c }$

Initial temperature of water $T_{w}$ = 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

⇒  $m_{a}$ $C_{a}$ ( $T_{a}$ - $T_{f}$ ) =  $m_{w}$ $C_{w}$ ( $T_{f}$ - $T_{w}$ ) ----------- (1)

Put all the values in equation (1)

⇒ 101 × 0.9 × (100.1 - $T_{f}$ ) = 48.9 × 4.2 × ( $T_{f}$ - 16.5 )

⇒ 90.9 × (100.1 - $T_{f}$ ) = 205.38 × ( $T_{f}$ - 16.5 )

⇒ 100.1 - $T_{f}$ = 2.26 ( $T_{f}$ - 16.5 )

⇒ 100.1 - $T_{f}$ = 2.26 $T_{f}$  - 37.28

⇒ 3.26 $T_{f}$ = 137.38

⇒ $T_{f}$ = 42.14 °C

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