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

The final temperature of the mixture is = 42.14 °C

Step-by-step explanation:

Mass of the aluminium
`m_(a)` = 101 gm

Specific heat for aluminium
`C_(a)` = 0.9
`(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
`(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