Question

A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.10 ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C)

Answer 1

Answer:

$m_{steel}=54.47gSteel$

Explanation:

Hello,

In this case, one considers that the heat lost by the water is gained by the steel rod:

$\Delta H_{water}=-\Delta H_{steel}$

That in terms of masses, heat capacities and temperatures is:

$m_{water}Cp_{water}(T_F-T_{water})=-m_{steel}Cp_{steel}(T_F-T_{steel})$

Thus, solving for the mass of the steel rod, we obtain:

$m_{steel}=\frac{m_{water}Cp_{water}(T_F-T_{water})}{-Cp_{steel}(T_F-T_{steel})} \\\\m_{steel}=\frac{125mL*\frac{1g}{1mL}*4.18\frac{J}{g^oC} (22.00^oC-21.10^oC)}{-0.452\frac{J}{g^oC}*(2.00^oC-21.10^oC)} \\\\m_{steel}=54.47gSteel$

Best regards.