Question

A combination of 0.250 kg of water at 20.0°C, 0.400 kg of aluminum at 26.0°C, and 0.100 kg of copper at 100°C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container and determine the final temperature of the mixture.

Answer 1

23.63 °C

Step-by-step explanation:

`m_(w)` = mass of water = 0.250 kg

`T_(wi)` = initial temperature of water = 20.0 °C

`c_(w)` = Specific heat of water = 4186 J/(kg °C)

`m_(a)` = mass of aluminum = 0.400 kg

`T_(ai)` = initial temperature of aluminum = 26.0 °C

`c_(a)` = Specific heat of aluminum = 900 J/(kg °C)

`m_(c)` = mass of copper = 0.100 kg

`T_(ci)` = initial temperature of copper = 100 °C

`c_(c)` = Specific heat of copper = 386 J/(kg °C)

`T_(f)` = Final temperature of mixture

Using conservation of heat

`m_(w)`
`c_(w)` (
`T_(f)` -
`c_(w)`) +
`m_(a)`
`c_(a)` (
`T_(f)` -
`T_(ai)` ) +
`m_(c)`
`c_(c)` (
`T_(f)` -
`T_(ci)` ) = 0

(0.250) (4186) (
`T_(f)` - 20) + (0.400) (900) (
`T_(f)` - 26) + (0.100) (386) (
`T_(f)` - 100) = 0

`T_(f)` = 23.63 °C