A 200 ohm coil electric kettle is placed in a vessel containing a mixture of water and ice.

A 200 ohm coil electric kettle is placed in a vessel containing a mixture of water and ice. The voltage in the network is 220 V. The mass of ice in the mixture is 100 g. After 10 minutes the temperature of the water in the vessel became equal to 20 degrees Celsius. How much water was originally in the vessel? Heat loss is negligible

R = 220 ohms.

U = 220 V.

ml = 100 g = 0.1 kg.

T = 10 min = 600 s.

t1 = 0 ° C.

t2 = 20 ° C.

Cw = 4200 J / kg * ° C.

q = 3.4 * 10 ^ 5 J / kg.

mv -?

The amount of heat Qн, which is necessary for heating water with ice, is determined by the formula: Qн = Cw * mw * (t2 – t1) + Cw * ml * (t2 – t1) + q * ml.

The amount of heat Qk, which is released in the boiler, is determined by the Joule-Lenz law: Qk = U ^ 2 * T / R, where U is the current voltage, T is the time of passage of current in the boiler, R is the resistance of the boiler.

The amount of heat that is released in the boiler goes to heating water and melting ice: Qн = Qк.

Cw * mw * (t2 – t1) + Cw * ml * (t2 – t1) + q * ml = U ^ 2 * T / R.

Cw * mw * (t2 – t1) = U ^ 2 * T / R – Cw * ml * (t2 – t1) – q * ml.

mv = (U ^ 2 * T / R – Cw * ml * (t2 – t1) – q * ml) / Cw * (t2 – t1).

mv = ((220 V) ^ 2 * 600 s / 220 Ohm – 4200 J / kg * ° C * 0.1 kg * (20 ° C – 0 ° C) – 3.4 * 10 ^ 5 J / kg * 0.1 kg) / 4200 J / kg * ° C * (20 ° C – 0 ° C) = 1.06 kg.

Answer: initially there was mw = 1.06 kg of water in the vessel.