contestada

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.

Respuesta :

Answer:

23.63 °C

Explanation:

[tex]m_{w}[/tex] = mass of water = 0.250 kg

[tex]T_{wi}[/tex] = initial temperature of water = 20.0 °C

[tex]c_{w}[/tex] = Specific heat of water = 4186 J/(kg °C)

[tex]m_{a}[/tex] = mass of aluminum = 0.400 kg

[tex]T_{ai}[/tex] = initial temperature of aluminum = 26.0 °C

[tex]c_{a}[/tex] = Specific heat of aluminum = 900 J/(kg °C)

[tex]m_{c}[/tex] = mass of copper = 0.100 kg

[tex]T_{ci}[/tex] = initial temperature of copper = 100 °C

[tex]c_{c}[/tex] = Specific heat of copper = 386 J/(kg °C)

[tex]T_{f}[/tex] = Final temperature of mixture

Using conservation of heat

[tex]m_{w}[/tex] [tex]c_{w}[/tex] ([tex]T_{f}[/tex] - [tex]c_{w}[/tex]) + [tex]m_{a}[/tex] [tex]c_{a}[/tex] ([tex]T_{f}[/tex] - [tex]T_{ai}[/tex] ) + [tex]m_{c}[/tex] [tex]c_{c}[/tex] ([tex]T_{f}[/tex] - [tex]T_{ci}[/tex] ) = 0

(0.250) (4186) ([tex]T_{f}[/tex] - 20) + (0.400) (900) ([tex]T_{f}[/tex] - 26) + (0.100) (386) ([tex]T_{f}[/tex] - 100) = 0

[tex]T_{f}[/tex] = 23.63 °C

Otras preguntas