Spent steam from an electric generating plant leaves the turbines at 120.0 degrees Celsius and is cooled to 90.0 degrees Celsuius liquid water by water from a cooling tower in a heat exchanger. How much heat is removed by the cooling tower water for each kilogram of spent steam?

q1 = heat removed by moving steam from 120.0 degrees C to steam @ 100.0 degrees C.
q1 = mass x specific steam steam x (Tf-Ti).

q2 = heat removed by condensing steam @ 100.0 degrees C to liquid water at 100.0 degrees C.
q2 = mass x heat of condensation (same as heat of vaporization but different sign).

q3 = heat removed by moving liquid water from 100.0 degrees C to 90.0 degrees C.

Total heat removed is q1 + q2 + q3.
The problem doesn't exactly specify a mass but it asks for heat per kg so I would use 1 kg or 1000g for mass in each of the above equations.

q1 = 1000g x 2.097 kJ/gK x (100.0 - 120.0) = -209.7 kJ

q2 = 1000g x 2257 kJ/kg = 2257 kJ
q3 = 1000g x 4.184 kJ/gK x (90.0 - 100.0) = -418.4 kJ

Total heat removed = q1 + q2 + q3 = 2257 - 209.7 - 418.4 = 1629.3 kJ/kg

To find the heat removed by the cooling tower water for each kilogram of spent steam, we can calculate the individual heat transfers using the given temperatures and properties. Let's calculate each step:

Step 1: Heat removed by moving steam from 120.0 degrees C to steam @ 100.0 degrees C (q1):
Using the equation q = mass x specific heat x (Tf - Ti), where mass = 1 kg and Tf = 100.0 degrees C, and Ti = 120.0 degrees C.

q1 = 1 kg x specific heat of steam x (100.0 degrees C - 120.0 degrees C)

Step 2: Heat removed by condensing steam @ 100.0 degrees C to liquid water at 100.0 degrees C (q2):
Using the equation q = mass x heat of condensation, where mass = 1 kg.

q2 = 1 kg x heat of condensation of steam

Step 3: Heat removed by moving liquid water from 100.0 degrees C to 90.0 degrees C (q3):
Using the equation q = mass x specific heat x (Tf - Ti), where mass = 1 kg, Tf = 90.0 degrees C, and Ti = 100.0 degrees C.

q3 = 1 kg x specific heat of water x (90.0 degrees C - 100.0 degrees C)

Finally, we can calculate the total heat removed by summing q1, q2, and q3:

Total heat removed = q1 + q2 + q3

Please note that for the specific heat capacity and heat of condensation, you may need to refer to a table or use known values specific to the substance being used in the electric generating plant.

To find the amount of heat removed by the cooling tower water for each kilogram of spent steam, we need to calculate q1, q2, and q3 and then sum them up.

q1 represents the heat removed by moving steam from 120.0 degrees Celsius to steam at 100.0 degrees Celsius. We can use the equation q1 = mass x specific heat of steam x (Tf - Ti), where Tf is the final temperature (100.0 degrees Celsius) and Ti is the initial temperature (120.0 degrees Celsius). The specific heat of steam is typically around 2.03 kJ/kgK.

So, q1 = 1 kg x 2.03 kJ/kgK x (100.0 - 120.0) = -40.6 kJ (negative sign indicates heat removal).

q2 represents the heat removed by condensing the steam at 100.0 degrees Celsius to liquid water at the same temperature. The heat of condensation (or vaporization) for water is typically around 2260 kJ/kg.

Therefore, q2 = 1 kg x (-2260 kJ/kg) = -2260 kJ.

q3 represents the heat removed by moving liquid water from 100.0 degrees Celsius to 90.0 degrees Celsius. Again, we can use the equation q3 = mass x specific heat of water x (Tf - Ti). The specific heat of water is roughly 4.18 kJ/kgK.

So, q3 = 1 kg x 4.18 kJ/kgK x (90.0 - 100.0) = -41.8 kJ.

Finally, we can find the total heat removed by summing up q1, q2, and q3:

Total heat removed = q1 + q2 + q3 = -40.6 kJ + (-2260 kJ) + (-41.8 kJ) = -2342.4 kJ.

Hence, the cooling tower water removes 2342.4 kJ of heat when 1 kilogram of spent steam is cooled from 120.0 degrees Celsius to 90.0 degrees Celsius.