# Two spirals of an electric stove with a resistance of 10 ohms each are connected in parallel and connected

**Two spirals of an electric stove with a resistance of 10 ohms each are connected in parallel and connected to a network with a voltage of 220 V. Water weighing 1 kg, poured into an aluminum pan weighing 300 g, boiled after 37 seconds. What is the starting temperature of the water and the pot? Disregard the energy losses for heating the ambient air. The specific heat capacity of water is 4200 J / (kg⋅ ° С), aluminum is 900 J / (kg⋅ ° С).**

R1 = R2 = 10 ohms.

U = 220 V.

m1 = 1 kg.

m2 = 300 g = 0.3 kg.

T = 37 s.

t2 = 100 ° C.

C1 = 4200 J / kg * ° С.

C2 = 900 J / kg * ° C.

t1 -?

With a parallel connection of the spirals, their total resistance R is found by the formula: R = R1 * R2 / (R1 + R2).

R = 10 ohms * 10 ohms / (10 ohms + 10 ohms) = 5 ohms.

According to the Joule-Lenz law, the amount of heat Q = U2 * T / R is released in the tile.

This amount of heat Q is used to heat water from t1 to the boiling point t2.

Q = C1 * m1 * (t2 – t1) + C2 * m2 * (t2 – t1) = C1 * m1 * t2 – C1 * m1 * t1 + C2 * m2 * t2 – C2 * m2 * t1.

U2 * T / R = C1 * m1 * t2 – C1 * m1 * t1 + C2 * m2 * t2 – C2 * m2 * t1.

C1 * m1 * t1 + C2 * m2 * t1 = C1 * m1 * t2 + C2 * m2 * t2 – U2 * T / R.

t1 = (C1 * m1 * t2 + C2 * m2 * t2 – U2 * T / R) / (C1 * m1 + C2 * m2).

t1 = (4200 J / kg * ° C * 1 kg * 100 ° C + 900 J / kg * ° C * 0.3 kg * 100 ° C – (220 V) 2 * 37 s / 5 Ohm) / (4200 J / kg * ° C * 1 kg + 900 J / kg * ° C * 0.3 kg) = 19.8 ° C.

Answer: the initial water temperature was t1 = 19.8 ° C.