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A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. After the sizzling subsides, what is the final equilibrium temperature? (Make the reasonable assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don’t affect the outcome, as we’ll see in the next section.)

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mairarmol

Answer:

[tex]T_{2}=16,97^{\circ}C[/tex]

Explanation:

The specific heats of water and steel are  

[tex]Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}[/tex]

[tex]Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}[/tex]

Assuming that the water and steel are into an adiabatic calorimeter (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium [tex]T_{2}_{w}= T_{2}_{s}[/tex]  

An energy balance can be written as

[tex] m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s} [/tex]  

Replacing

[tex] 1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)[/tex]

Then, the temperature [tex]T_{2}=16,97^{\circ}C[/tex]

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