# Problem No. 8 (10 points)

A water heater with a power of \( P = 1000 \mathrm{W} \) is used to heat a certain amount of water. When the heater is turned on for \( t_{1} = 2 \) minutes, 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 \( t_{2} = 1 \) minute. 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 \(\cdot{ }^{\circ} \mathrm{C}\)

Answer: 4.76 kg
# Solution and Evaluation Criteria:

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

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

As a result, we get: $m=\frac{P \cdot t_{1} \cdot t_{2}}{c \Delta T\left(t_{1}+t_{2}\right)}=\frac{1000 \cdot 120 \cdot 60}{4200 \cdot 2 \cdot(120+60)}=4.76 \mathrm{kg}$

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-5.jpg?height=354&width=352&top_left_y=197&top_left_x=161)

Multidisciplinary Engineering Olympiad "Star" in Natural Sciences

## Final Stage

2016-2017 academic year

Tasks, answers, and evaluation criteria

![](https://cdn.mathpix.com/cropped/2024_05_06_3676ffec89688f4f2c70g-5.jpg?height=123&width=554&top_left_y=641&top_left_x=908)

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