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A 25.0−gsample of an alloy at 93.00°Cis placed into 50.0 gof water at 22.00°Cin an insulated coffee-cup calorimeter with a heat capacity of 9.20 J/K.If the final temperature of the system is 31.10°C,what is the specific heat capacity of the alloy?

Respuesta :

Answer:

1.23 J/(g °C)

Explanation:

[tex]m_{w}[/tex] = mass of water = 50 g

[tex]c_{w}[/tex] = specific heat of water = 4.186 J/(g °C)

[tex]T_{wo}[/tex] = Initial temperature of water = 22.00 °C

[tex]m_{a}[/tex] = mass of alloy = 25 g

[tex]c_{a}[/tex] = specific heat of alloy = ?

[tex]T_{ao}[/tex] = Initial temperature of alloy = 93.00 °C

[tex]T_{f}[/tex] = Final equilibrium temperature = 31.10 °C

Using conservation of heat

Heat Lost by alloy = Heat gained by water

[tex]m_{a}c_{a}(T_{ao}-T_{f})[/tex] = [tex]m_{a}c_{a}(T_{f} - T_{ao})[/tex]

(25) [tex]c_{a}[/tex] (93 - 31.10) = (50) (4.186) (31.10 - 22)

[tex]c_{a}[/tex] = 1.23 J/(g °C)

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