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

Answer:

23.63 °C

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