A rigid tank is divided into two equal parts by a partition. Initially one side of the tank contains 5 kg of water at 200 kPa and 25 C, and the other side is evacuated (i. e. there is nothing there, not even air). The partition is then removed and the water expands into the entire tank. The water is allowed to exchange heat with its surroundings until the temperature in the tank returns to the initial value of 25 C. Determine the heat transfer into the water for this process in kJ.

There is no heat transfer in the system as the temperature before and after remains same i.e, 25 C.

Explanation:

Heat energy is mathematically expressed as,

     Q=mCΔT

By looking at the above equation we can say that heat gain or loss by any system is equal to the mass times specific heat multiplied by the change in temperature.  

Now if we consider the water system in view of this equation then ΔT would be 0 therefore it is evident that there is no transfer of heat.

If we consider the empty half of the tank so it is mentioned that there is nothing no air and giving us m=0, hence Q=0.

Mathematically we can express this as

Q=mCΔT

Q=(5)(4200)(0)

Q= 0 KJ (which indicates that there was no heat gain or loss by the system)

d.7.5cm is your answer

v=ir = 3x15 =45 v