[relativistic]

OK , I HAVE THE WAY OF SOLUTION , I HOPE YOU UNDERSTAND ENGLISH

SOLVING PROBLEMS LIKE THIS NEED COMBINING ELECTRICITY WITH HEAT.

THE KEY POINT HERE IS

ELECTRICAL ENERGY LOST BY HEATER {ELECTRIC WORK DONE TO MOVE THE CHARGES THROUGH THE WIRE} EQUALS THE HEAT ENERGY GAINDE BY WATER.

WE HAVE TO CALCULATE THE ELECTRICAL ENERGY{W}
V = W / Q
W = V * Q { V IS THE POTENTIAL DIFFRENCE , Q IS THE AMOUNT OF CHARGES }

WE HAVE TO KNOW THE AMOUNT OF CHARGES {Q}

WE HAVE THE LAW
I = Q/t SO
Q = I * t {I IS CURRENT INTENSTY WHILE t IS TIME IN SECONDS}
WE CAN KNOW I
I = V / R {R IS THE RESISTANCE}

I = 220/55 =4 AMPERE
SO
Q= 4 *42 * 60=10080 COLUMB
SO
W = 220 * 10080 = 2217600 JOULE { THE FIRST PART}

WE HAVE ELECTRICAL ENERGY EQUAL HEAT ENERGY
HEAT ENERGY = m * c * change in tempreature
{m is mass , c is specific heat capacity}
so
change in tempreature = W/(m*c)
m= 10 KG , c= 4200 JOULE/ (KG * KELVIN) , W = HEAT ENERGY OR ELECTRICAL ENERGY
SO WATER TEMPREATURE RISES BY {52.8 DEGREE KELVIN OR DEGREE CELSIUS}