What happens to the outlet temperatures of the hot and cold fluid when the both the hot fluid is at a maximum and the cold fluid is at a minimum flowrate?
a- remains same as that of inlet temperature
b- hot water temperature decreases and cold water temperature increase
c- hot water remains almost same but cold water temperature increases drastically.

c- Hot water remains almost same but cold water temperature increases drastically.

Explanation:

In a process of heat exchange where a hot and a cold fluid are present, the amount of heat that is transfered from one to another, the heat lost by the hot fluid, is gained by the cold fluid. That can be calculated by the equation : Q = m * Cp * dT.

Where:

m= mass flowrate

Cp=specific-heat-capacity

dT= difference between the inlet and outlet temperatures

We can state the equality of this equation for both fluids:

(mass-flowrate * specific-heat-capacity * (temperature-in – temperature-out) ) for hot medium = (mass-flowrate * specific-heat-capacity * (temperature-out – temperature-in) ) for cold medium

Now if we increase the value of the mass flow rate of the hot fluid to its maximum and decrease the mass flowrate of the cold fluid to its minimum. That will mean that the difference between the inlet and outlet temperatures for the cold fluid must increase so as to compensate the increase on the other side of the equation and guarantee the equality. And taking in account that the inlet temperature of the cold fluid is known and what can change is the outlet one, the correct answer is c that says that the outlet temperature of the cold fluid increases drasically.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j