A1.28-kg sample of water at 10.0 °c is in a calorimeter. you drop a piece of steel with a mass of 0.385 kg at 215 °c into it. after the sizzling subsides, what is the final equilibrium temperature? (make the reasonable assumptions that any steam produced condenses into liquid water during the process of equilibration and that the evaporation and condensation don’t affect the outcome, as we’ll see in the next section.)

Explanation:

The specific heats of water and steel are  

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  

An energy balance can be written as

Replacing

Then, the temperature

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%