What will be the final temperature of the water in an insulated container as the result of passing 5.00 g of steam [H2O(g)] at 100.0°C into 195.0 g of water at 20.0°C? (?H°vap = 40.6 kJ/mol H2O)

The best way to do these problems is to make the heat gained + heat lost = 0.
q1 = heat gained by water = mcdeltaT
massH2O x specific heat water x (Tf-Ti). Tf is the final T and Ti is the initial T.
q1 = 195.0 g x 4.184 J/g*C x (Tf - 20.0)
q2 = heat lost by steam in converting steam (g) @ 100.0 C to liquid @ 100.0 C = mass x delta H.
q2 = 5.00g x 40,600 Joules = ??
q3 = heat lost with 5.00 g (from the steam conversion)water moving from 100.0 C to final T.
q3 = 5.00g x 4.184 J/g*C x (Tf - Ti).
Ti is 100.0.
q1 + q2 + q3 = 0.
Plug in for q1, q2, and q3 and solve for Tf.

Why did the water go to therapy? Because it had too many heat problems and needed some counseling!

To find the final temperature of the water, we need to set up the equation q1 + q2 + q3 = 0.

q1 represents the heat gained by the water, which is given by the formula:
q1 = massH2O x specific heat water x (Tf - Ti)

Given:
massH2O = 195.0 g
specific heat water = 4.184 J/g*C
Ti = 20.0°C

q1 = 195.0 g x 4.184 J/g*C x (Tf - 20.0)

q2 represents the heat lost by the steam in converting from steam to liquid at 100.0°C, which is given by the formula:
q2 = mass x delta H

Given:
mass = 5.00 g
delta H = 40,600 J/mol H2O

q2 = 5.00 g x 40,600 J/g = ??

q3 represents the heat lost by the steam in cooling from 100.0°C to the final temperature, which is given by the formula:
q3 = mass x specific heat water x (Tf - Ti)

Given:
mass = 5.00 g
specific heat water = 4.184 J/g*C
Ti = 100.0°C

q3 = 5.00 g x 4.184 J/g*C x (Tf - 100.0)

Now, we can substitute the values of q1, q2, and q3 into the equation q1 + q2 + q3 = 0 and solve for Tf to find the final temperature of the water.

To solve this problem, we need to follow the steps provided:

Step 1: Calculate the heat gained by the water (q1).
To do this, we use the formula q1 = mcdeltaT, where m is the mass of water, c is the specific heat capacity of water, and deltaT is the change in temperature.
q1 = 195.0 g x 4.184 J/g*C x (Tf - 20.0)

Step 2: Calculate the heat lost by the steam in converting from steam to liquid (q2).
The heat lost is given by q2 = mass x deltaH, where mass is the mass of steam and deltaH is the enthalpy of vaporization.
q2 = 5.00 g x 40,600 J/g

Step 3: Calculate the heat lost by the water from the steam conversion (q3).
We use the same formula as in step 1, but this time the initial temperature (Ti) is 100.0°C.
q3 = 5.00 g x 4.184 J/g*C x (Tf - 100.0)

Step 4: Set up the equation q1 + q2 + q3 = 0 and solve for Tf.
Plug in the values for q1, q2, and q3 from steps 1-3 into the equation and solve for Tf.

Note: The equations used assume that the specific heat capacity and enthalpy of vaporization do not change with temperature.

By substituting the given values into the equations and solving, we can find the final temperature (Tf) of the water in the insulated container.

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