A0.95-kg sample of water at 11°c is in a calorimeter. you drop a piece of steel with a mass of 0.39 kg at 212°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 taken together don't affect the outcome, as we'll see later.

20 °C

Explanation:

The calorimeter is an isolated system, so no heat is lost to the surroundings:

Qnet = 0

We will designate the steel (assumed to be carbon steel) as system 1 and the water as system 2, so the equation becomes:

Q₁ + Q₂ = 0

m₁c₁(Teq - Ti₁) + m₂c₂(Teq - Ti₂) = 0

As indicated in the problem, we will assume the steel only heats the water to raise its temperature and does not vaporize the water.

We substitute in the values and solve for Teq, the final equilibrium temperature. The specific heat capacity of water is 4184 Jkg⁻¹°C⁻¹ and of carbon steel, 502.416 Jkg⁻¹°C⁻¹:

(0.39 kg)(502.416 Jkg⁻¹°C⁻¹)(Teq - 212 °C) + (0.95 kg)(4184 Jkg⁻¹°C⁻¹)(Teq - 11 °C) = 0

(195.94224 J°C⁻¹)(Teq - 212 °C) + (3974.8 J°C⁻¹)(Teq - 11 °C) = 0

(195.94224 J°C⁻¹)Teq - 41539.75 J + (3974.8 J°C⁻¹)Teq - 43722.8 J = 0

(4170.74224 J°C⁻¹)Teq = 85262.55 J

Teq = 20 °C

divide the right side by the left side

0.024÷1.2=0.02 for example

this could be newton#'s law of cooling. the cooling curve is exponential