Question 428605
This would seem to be a physics problem rather than a math problem. But the solution is as follows:
The energy gained by the water due to heating = mass of the water * specific heat * temperature rise:
E = mc{{{DELTA}}}T
The power dissipated in the resistor is given up as heat in the water.
{{{P = V^2/R = 230^2/25 = 2116}}}.  The power units are J/s.  Therefore, the total energy given to the water in the form of heat in t seconds is:
2116 J/s * t
So we have 
2116 J/s * t  = 0.85 kg * 4190 J/k/deg-C * 20 deg-C
Solve for t:
t = 4190*20*0.85/2116 = 33.66 sec.