It’s been a while since I did these sorts of sums but..
For the temperature in the tank to be constant the power entering must match that extracted so first work out the power in Watts (=Joules/second) entering the tank via the hot coil using:
Heat capacity of water = 4180 J/Kg/K
Temperature drop (10 K)
Mass of water per second (3kg/S)
= 4180 * 10 * 3
= 125,400 J/S
= 125 kW
So that's how much has to be removed by the cold water flow.
Quick check that the coil is big enough. Last time I did this I used this equation:
Dt = (P * Wt)/(A * C)
Where
Dt = average delta T across wall of pipe in Centigrade
C = thermal conductivity of copper = 401 W•m−1•K−1
Wt = wall thickness in meters? Assume 2mm = 0.002m ?
A = area of pipe wall in meters = Pi * 0,02 * 30 = 1.88sqm
P = Power in watts
Dt=(125*10^3 * 0.002)/(1.88 * 401)
=0.33K
So can assume no temperature drop across wall of copper pipe (unless it scales up!)
Incoming hot is 50C, require 10C drop so it leaves at 40C. Average 45C
Lets assume water temperature of the tank is maintained at 45C.
Temperature rise of cooling water is 45-20C = 25C
Flow rate required is
= 125,400/(4180 * 25)
= 1.2L/Sec
Note that it will come out colder than required until the tank is upto temperature. that might take quite awhile.