120 g of water is poured into an aluminum calorimeter weighing 50 g and a 2-ohm spiral connected to a 15-volt

120 g of water is poured into an aluminum calorimeter weighing 50 g and a 2-ohm spiral connected to a 15-volt voltage source is lowered, how long will the calorimeter with water heat up by 9 degrees if energy losses to the environment can be neglected.

ma = 50 g = 0.05 kg.

Ca = 920 J / kg * ° C.

mw = 120 g = 0.12 kg.

Cw = 4200 J / kg * ° C.

R = 2 ohms.

U = 15 V.

ΔT = 9 ° C.

t -?

Heat energy Q is released in the conductor, the value of which is determined by the formula: Q = U ^ 2 * t / R.

This heat Q goes to heating the water and the calorimeter: Q = Ca * ma * ΔT + Cw * mw * ΔT = (Ca * ma + Cw * mw) * ΔT.

U ^ 2 * t / R = (Ca * ma + Cv * mv) * ΔT.

t = (Ca * ma + Cw * mv) * ΔT * R / U ^ 2.

t = (920 J / kg * ° C * 0.05 kg + 4200 J / kg * ° C * 0.12 kg) * 9 ° C * 2 Ohm / (15 V) ^ 2 = 44 s.

Answer: it takes time t = 44 s to heat water.