Question

Термос емкостью V=3л заполнили кипящей водой и закрыли. Через двое суток температура воды стала t(к)=70'C. Какое количество теплоты отдала вода за это время?

Answer 1

Ответ:

378 мДж количество теплоты отдала вода за это время

Объяснение:

$V=3l=3*10^{-3} m^3$

температура кипения воды равна 100°C (лучше запомнить)

$t_0=$ 100°C

$t=$ 70 °C

Удельная теплоёмкость воды с=4200 Дж/(кг*°С) (табличное значение)

$Q=cm(t-t_0)$

m=pV, p = 1000 $m^3$ (табличное значение)

m=$3*10^{-3}*1000=3$ кг

$Q=4200*3*(70-100)=12600*(-30)=-378000$ Дж

$-378000$ Дж = - $378 * 10^{-3}$ Дж= -378 мДж

Answer 2

Calculating Heat Released by Water in a Thermos

To calculate the amount of heat released by the water in the thermos, we can use the formula:

Q = mcΔT

Where: - Q = heat released (in joules) - m = mass of the water (in grams) - c = specific heat capacity of water (in J/g°C) - ΔT = change in temperature (in °C)

First, let's calculate the mass of the water in the thermos. We know the volume of the water is 3 liters, and the density of water is approximately 1 gram per milliliter. Therefore, the mass of the water can be calculated as:

m = V × ρ

Where: - V = volume of water (in liters) - ρ = density of water (in g/mL)

Substituting the given values: m = 3 L × 1000 g/L = 3000 g

Next, we need to find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.

Now, we can calculate the heat released by the water using the given change in temperature: ΔT = t(k) - t(initial) = 70°C - 100°C = -30°C

Substituting the values into the formula: Q = 3000 g × 4.18 J/g°C × -30°C = -376,200 J

Therefore, the water released approximately 376,200 joules of heat over the two-day period.