How long will it take for water with a mass of 200 grams to boil if the current in the boiler is I = 0.5 A

How long will it take for water with a mass of 200 grams to boil if the current in the boiler is I = 0.5 A at a network voltage of U = 220 V? The initial water temperature is t = 20 degrees C.

We give all the values ​​from given in the SI system:
m = 200 g = 0.2 kg.
t1 = 20 ° C = 293 K.
t2 = 100 ° C = 373 K.
1. The power of the heater is determined by the formula:
P = Q / τ, Q – where the amount of heat, τ – heating time.
Hence Q = P * τ
2. The amount of heat spent on heating the body is equal to the product of the specific heat capacity of the substance, body weight and the difference between the final and initial temperatures.
Q = c * m * (t2-t1), where c is the specific heat capacity of the substance, m is the mass of the substance, t2 and t1 are the final and initial temperatures, respectively.
Specific heat capacity of water c = 4200 J / kg * K,
3. Formula for determining electrical power:
P = U * I
4. Substitute the expressions from points 2 and 3 into the expression from point 1:
c * m * (t2-t1) = U * I * τ
Let us express time from this expression:
τ = (c * m * (t2-t1)) / (U * I)
5. Substitute the numbers and determine the time until boiling:
τ = (c * m * (t2-t1)) / (U * I) = 4200 * 0.2 * (373-293) / (220 * 0.5) = 610 s.
Answer: the water will boil in 610 s.