The mug contains a mixture of water and ice. After the contents of the mug had been heated with a boiler
The mug contains a mixture of water and ice. After the contents of the mug had been heated with a boiler for 2 minutes, it contained 300 ml of water at a temperature of 30 ° C. The boiler is powered by the mains.
Let’s designate the data: T1 = 0, t = 2 min = 120 s, v = 300 ml = 0.3 L = 0.0003 m ^ 3, T2 = 30 ° C, U = 220 V, R = 95 Ohm, m – ice mass, M – water mass, j = 335.5 * 10 ^ 3 J / kg, p = 1000 kg / m ^ 3, c = 4200 J / (kg * ° C), m -?
In order for the ice to melt, it is necessary to transfer heat to it, equal to Q1 = j * m
In order for the water to heat up to a temperature of 30 ° C, it is necessary to transfer heat to it, equal to Q2 = c * M * T2 = c * p * v * T2
In 120 seconds, they completed work equal to A = P * t = t * U ^ 2 / R
According to the law of conservation of energy, Q1 + Q2 = A => j * m + c * p * v * T2 = t * U ^ 2 / R
m = (t * U ^ 2 / R – c * p * v * T2) / j = (120 * 220 ^ 2/95 – 4200 * 1000 * 0.0003 * 30) / (335.5 * 10 ^ 3 ) = 69 * 10 ^ -3 kg = 69 g