Answer :
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]
The equilibrium temperature given that You drop a piece of steel at 215 °C into the water at 10.0 °C in the calorimeter is 16 °C
How to calculate the equilibrium temperature?
The equilibrium temperature given that You drop a piece of steel at 215 °C into the water at 10.0 °C can be calculated as follow:
- Mass of steel (M) = 0.385 Kg
- Temperature of steel (T) = 215 °C
- Specific heat capacity of steel (C) = 420 J/kg°C
- Mass of water (Mᵥᵥ) = 1.28 Kg
- Temperature of water (Tᵥᵥ) = 10.0 °C
- Specific heat capacity of water (Cᵥᵥ) = 4184 J/Kg°C
- Equilibrium temperature of steel and water mixture (Tₑ) =?
[tex]MC(T - T_e) = M_wC_w(T_e - T_w)\\\\0.385\ \times\ 420(215 - T_e) = 1.28\ \times\ 4184(T_e - 10)[/tex]
[tex]34765.5 - 161.7T_e = 5355.52T_e - 53555.2\\\\34765.5\ +\ 53555.2 = 5355.52T_e\ +\ 161.7T_e\\\\88320.7 = 5517.22T_e\\\\T_e = \frac{88320.7}{5517.22} \\\\T_e = 16\ \textdegree C[/tex]
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