Question

Из чайника выкипела вода объемом 0,5 л, начальная температура которой была равна 10 °С. Какое количество теплоты оказалось излишне затраченным?

Answer 1

Q=2,3*10^6 Дж/кг*1*10^3 кг/м^3*0.5*10^-3 м^3=1.15*10^6 Дж=1.15Мдж

Answer 2

Calculating Excess Heat Loss from the Kettle

To calculate the excess heat loss from the kettle, we can use the formula:

Q = mcΔT

Where: - Q is the heat energy transferred (in joules) - m is the mass of the substance (in kilograms) - c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius) - ΔT is the change in temperature (in degrees Celsius)

Given: - Initial temperature, T1 = 10 °C - Final temperature, T2 = 100 °C (assuming the water reached boiling point) - Mass, m = 0.5 kg (since the density of water is approximately 1 kg/L)

We can use the specific heat capacity of water, which is approximately 4186 J/kg°C.

Using the formula, we can calculate the heat energy transferred:

Q = mcΔT = 0.5 kg * 4186 J/kg°C * (100 °C - 10 °C)

Q = 0.5 kg * 4186 J/kg°C * 90 °C

Q = 188,370 J

So, the amount of heat energy transferred to the water to raise its temperature from 10 °C to 100 °C is approximately 188,370 joules.

However, the excess heat loss can be calculated by subtracting the actual heat energy transferred from the ideal heat energy transferred. The ideal heat energy transferred can be calculated using the final temperature of the water and the initial temperature of the water.

Ideal Q = mcΔT = 0.5 kg * 4186 J/kg°C * (100 °C - 10 °C)

Ideal Q = 188,370 J

Therefore, the excess heat loss is 0 joules, as the actual heat energy transferred is equal to the ideal heat energy transferred.