Answer:
[tex]\Delta H=1149.5J[/tex]
Explanation:
Hello!
In this case, since the calorimeter contains a solution for which the specific heat is 4.18 J/g°C, which is equal to that of the water, we can also notice that the density is 1.00 g/mL and therefore the present mass is 250 g. In such a way, we compute the enthalpy change as shown below:
[tex]\Delta H=mC\Delta T[/tex]
Thus, by plugging in the mass, specific heat and temperature change, we obtain:
[tex]\Delta H=250g*4.18\frac{J}{g\°C}(26.1\°C-25.0\°C)\\\\\Delta H=1149.5J[/tex]
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