In an aluminum pan weighing 800 g, into which 2 kg of water is poured, an electric heater with a resistance

In an aluminum pan weighing 800 g, into which 2 kg of water is poured, an electric heater with a resistance of 50 ohms is lowered, through which a current of 4.5 A. passes. How many degrees will the water in the pan heat up in 10 minutes if the heat loss is 15%? Specific heat of aluminum 880 J / kg * K; water-4200 J / kg * K

Given:
m1 = 800g = 0.8kg,
m2 = 2kg,
c1 = 880J / (kg * deg),
c2 = 4200J / (kg * deg),
t = 10min = 600s,
R = 50 Ohm,
I = 4.5A,
η = 15%;
Find: Δt -?
From the Joule-Lenz law, we find the amount of heat that the electric heater emits:
Q = I ^ 2 * R * t = 20.25 * 50 * 600 = 607500J;
the amount of heat spent on heating the water and the pan:
Qp = (1 – η) * Q / 100% = 516375J;
Qп = Q1 + Q2,
where
Q1 = m1 * c1 * Δt – the amount of heat for heating the pan,
Q2 = m2 * c2 * Δt – the amount of heat for heating water;
From here we find:
Δt * (m1 * c1 + m2 * c2) = 516375;
Δt = 516375 / (704 + 8400) = 56.72 ° C.