Answer:
[tex]\Delta H=687.4 J[/tex]
Explanation:
Hello!
In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:
1. Heat up from 298.15 K to 1673 K.
2. Undergo the phase transition.
Both process have an associated enthalpy as shown below:
[tex]\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J[/tex]
[tex]\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J[/tex]
Therefore, the required heat is:
[tex]\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J[/tex]
Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.
Best regards!
