# Problem No. 8 (10 points)

A water heater with a power of \( P = 500 \mathrm{W} \) is used to heat a certain amount of water. When the heater is turned on for \( t_{1} = 1 \) minute, the temperature of the water increases by \( \Delta T = 2^{\circ} \mathrm{C} \), and after the heater is turned off, the temperature decreases back to the initial value in \( t_{2} = 2 \) minutes. Determine the mass of the heated water, assuming the heat loss power is constant. The specific heat capacity of water \( c_{B} = 4200 \) J/kg \(\cdot{ }^{\circ} \mathrm{C}\)
# Solution and evaluation criteria:

The law of conservation of energy during the heating of water: $P \cdot t_{1}=c m \Delta T+P_{\text {loss }} \cdot t_{1}$.

When the heating plate is turned off: $P_{\text {loss }}=\frac{c m \Delta T}{t_{2}}$.

As a result, we get: $m=\frac{P \cdot t_{1} \cdot t_{2}}{c \Delta T\left(t_{1}+t_{2}\right)}=\frac{500 \cdot 120 \cdot 60}{4200 \cdot 2 \cdot(120+60)}=2.38$ kg