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

`m_(steel)=54.47gSteel`

Step-by-step 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)=(m_(water)Cp_(water)(T_F-T_(water)))/(-Cp_(steel)(T_F-T_(steel))) \\\\m_(steel)=(125mL*(1g)/(1mL)*4.18(J)/(g^oC) (22.00^oC-21.10^oC))/(-0.452(J)/(g^oC)*(2.00^oC-21.10^oC)) \\\\m_(steel)=54.47gSteel`

Best regards.