A 1.28kg sample of water at 10.0C is in a calorimeter. You drop a piece of steel with a mass of 0.385kg at 215C 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.)

The answer to the question is 16.04 °C

Explanation:

Mass of water mw = 1.28 kg

Mass of steel = 0.385 kg

Temperature of water = 10 °C

Temperature of steel = 215 °C

Specific heat capacity of water = 4.18 J/°C g

Specific heat capacity of steel = 420J/°C kg

The first law of thermodynamics states that energy is neither created nor destroyed but changes from one form to another

Therefore heat lost by steel = heat gained by water

mw×cw×(T-Tw) = ms×cs×(Ts-T)

1.28×4180×(T-10) = 0.385×420×(215-T)

Solving the above equation, we have T = 16.04 °C

the final equilibrium temperature = 16.04 °C