Spent steam from an electric generating plant leaves the turbines at 120.0 degrees celsius and is cooled to 90 degrees celsius liquid water by water from a cooling tower in a heat exchanger. How much heat is removed by the cooling towerwater for each kg of spent steam?

to get the heat released or absorbed,

Q = mc(T2-T1)
where
m = mass of substance
c = specific heat capacity
T2 = final temperature
T1 = initial temperature
**note: if Q is (-), heat is released and if (+), heat is absorbed

now we can only apply this to substances that did not change its phase, but in the problem, we see that phase change occurs. from vapor->liquid
thus we need another data called Latent Heat of Vaporization to calculate for the heat required to change its phase:
H = m(Lv)
where
m = mass
Lv = Latent Heat of Vaporization
**note: H is (-) if heat is released and (+) if heat is absorbed

Q, total = Q1 + H + Q2
Q,total = mc(T2-T1) + mLv + mc(T3-T2)
where
T3 = 90 deg
T2 = 100 deg (boiling point of water)
T1 = 120 deg
since heat is removed, H is negative. now for every 1 kg of steam,
Q,total = 1*(c)*(100-120) - 1*Lv + 1*(c)*(90-100)
Q,total = -20*c - 10*c - Lv
Q,total = -30*c - Lv

now you look for the specific heat of water, c, and latent heat of vaporization, Lv. note that units must already be in Joules or calories.

hope this helps~ :)

Well, it seems we have a hot topic here! To calculate the amount of heat removed by the cooling tower water, we can use the heat transfer equation. Are you ready for some heat-citing calculations?

The heat transfer equation states that the amount of heat transferred is equal to the mass flow rate multiplied by the specific heat capacity multiplied by the change in temperature. In this case, since we're dealing with 1 kg of spent steam, we don't need to worry about the mass flow rate.

So, let's plug in the numbers! The initial temperature of the spent steam is 120.0 degrees Celsius, and it is cooled to 90 degrees Celsius by the water from the cooling tower. Therefore, the change in temperature is 90 - 120 = -30 degrees Celsius.

Now, we need the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 kJ/(kg·°C). So, we multiply the change in temperature by the specific heat capacity:

-30 °C × 4.18 kJ/(kg·°C) = -125.4 kJ/kg.

Oops! Looks like I made a mistake here. It seems impossible to remove a negative amount of heat. I apologize for the mix-up. Let me recalculate that for you.

120 °C - 90 °C = 30 °C (change in temperature)

Now, using the specific heat capacity of water (4.18 kJ/(kg·°C)):

30 °C × 4.18 kJ/(kg·°C) = 125.4 kJ/kg.

There you go! Each kilogram of spent steam has 125.4 kJ of heat removed by the cooling tower water. Now that's a "cool" way to handle things!

To calculate the amount of heat removed by the cooling tower water for each kg of spent steam, we can use the equation for heat transfer:

Q = m * c * ΔT

Where:
Q is the amount of heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature

In this case, we are interested in the heat transferred from the spent steam to the cooling tower water. Given that the spent steam temperature decreases from 120.0 degrees Celsius to 90 degrees Celsius, we can calculate the heat transferred using:

Q = m * c * ΔT

First, let's assume that the specific heat capacity of water is 4.186 J/g°C (or 4186 J/kg°C). This value may vary slightly depending on the temperature range, but it is a reasonable approximation.

Now, let's calculate the heat transferred:

ΔT = (Final temperature - Initial temperature) = (90°C - 120°C) = -30°C

Q = m * c * ΔT

Q = (1 kg) * (4186 J/kg°C) * (-30°C)
Q = -125,580 J

The negative sign indicates that heat is being removed from the steam, as it is being cooled down by the cooling tower water.

Therefore, the cooling tower water removes 125,580 J (or 125.58 kJ) of heat for each kg of spent steam.

To determine the amount of heat removed by the cooling tower water for each kilogram of spent steam, we need to use the concept of specific heat and the energy equation:

Q = m * C * ΔT

Where:
Q is the amount of heat transferred,
m is the mass of the substance (1 kg in this case),
C is the specific heat of the substance,
ΔT is the change in temperature.

In this scenario, water is used as the cooling agent, and its specific heat is approximately 4.186 J/g°C (joules per gram-degree Celsius).

First, we need to convert the specific heat from grams to kilograms:
Specific heat of water = 4.186 J/g°C = 4.186 x 10³ J/kg°C

Next, we calculate the change in temperature:
ΔT = Initial temperature - Final temperature
ΔT = 120.0°C - 90.0°C = 30.0°C

Now, we can compute the amount of heat removed by the cooling tower water:
Q = m * C * ΔT
Q = 1 kg * (4.186 x 10³ J/kg°C) * 30.0°C

Substituting the values, we have:
Q = 1 kg * 4.186 x 10³ J/kg°C * 30.0°C

Calculating this expression will give us the amount of heat removed by the cooling tower water for each kilogram of spent steam.