Suppose 0.0150 kg of steam (at 100.00°C) is added to 0.150 kg of water (initially at 19.3°C.). The water is inside a copper cup of mass 54.1 g. The cup is inside a perfectly insulated calorimetry container that prevents heat flow with the outside environment. Find the final temperature (in °C) of the water after equilibrium is reached.

The heat gained by the liquid water and the copper cup while they reach a final equilibrium temperature Tf equals the heat lost by the steam in the process of condensing and cooling to the same temperature.

Write that as an equation and solve for Tf. Don't forget to include the heat of condensation, 540 calories per gram

To find the final temperature of the water after equilibrium is reached, we need to consider the heat gained by the cold water, the heat lost by the hot steam, and the heat gained by the copper cup.

First, let's calculate the heat gained by the cold water, using the specific heat capacity formula:

Q1 = m1 * c1 * ΔT1

Where:
Q1 = heat gained by the cold water
m1 = mass of the water
c1 = specific heat capacity of water (4.18 J/g°C)
ΔT1 = change in temperature of the water (final temperature - initial temperature)

m1 = 0.150 kg = 150 g
c1 = 4.18 J/g°C
ΔT1 = final temperature - initial temperature = Tf - 19.3°C

Next, let's calculate the heat lost by the hot steam, using the specific heat capacity and latent heat of vaporization formula:

Q2 = (m2 * c2 * ΔT2) + (m2 * Lv)

Where:
Q2 = heat lost by the hot steam
m2 = mass of the steam
c2 = specific heat capacity of steam (2.03 J/g°C)
ΔT2 = change in temperature of the steam (final temperature - initial temperature)
Lv = latent heat of vaporization of steam (2257 J/g)

m2 = 0.0150 kg = 15 g
c2 = 2.03 J/g°C
ΔT2 = final temperature - 100.0°C
Lv = 2257 J/g

Finally, let's calculate the heat gained by the copper cup, using the specific heat capacity formula:

Q3 = m3 * c3 * ΔT3

Where:
Q3 = heat gained by the copper cup
m3 = mass of the copper cup
c3 = specific heat capacity of copper (0.386 J/g°C)
ΔT3 = change in temperature of the copper cup (final temperature - initial temperature)

m3 = 54.1 g
c3 = 0.386 J/g°C
ΔT3 = final temperature - 19.3°C

Since the system is perfectly insulated, the heat gained by the cold water, heat lost by the hot steam, and heat gained by the copper cup must balance each other.

Q1 + Q3 = Q2

Now, we can set up the equation with the calculated values:

(m1 * c1 * ΔT1) + (m3 * c3 * ΔT3) = (m2 * c2 * ΔT2) + (m2 * Lv)

Substituting the values and solving for Tf (final temperature):

[(150 g) * (4.18 J/g°C) * (Tf - 19.3°C)] + [(54.1 g) * (0.386 J/g°C) * (Tf - 19.3°C)] =
[(15 g) * (2.03 J/g°C) * (Tf - 100.0°C)] + [(15 g) * (2257 J/g)]

Simplifying the equation would give the final temperature (Tf) as the solution.