Our power plant generates electricity using a steam turbine. At the exit from our power generation process, we end up with 562.0 kg/s of liquid water at 330.0 K. We cannot release this water directly into a nearby river if the temperature is over 310.00 K, without causing significant thermal pollution that would damage fish populations. Our plan is to mix this warm water with another cooler water stream at 290.00 K. What is the mass flow rate of this cool water stream needed to meet the regulation

The mass flowrate of the cooling water needed to meet the regulation is 560 kg/s

Explanation:

The given information are;

The mass flow rate of the water = 562.0 kg/s

The temperature of the exit water = 330.0 K

The allowable temperature of the of released water into nearby river = 310.00 K

The temperature of the cool water with which the temperature of the exit water is cooled = 290.00 K

Therefore, we have;

Heat gained by the mixture = The heat of the final mixture exited to the river

Q = m×c×ΔT

= 562.0 kg/s

The mass of the cooling water = x

The final temperature = 310 K

4.2×562×(330 - 310) = x×4.2×(310 - 290)

47208 = 84×x

x = 47208/84 = 560 kg

Therefore, the mass flowrate of the cooling water needed to meet the regulation = 560 kg/s.

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2