Question

ФИЗИКА ПОМОГИТЕ Температура воды в одном из горячих озёр Камчатки равна 48 °С. Сколько-литров этой воды нужно взять, чтобы, добавив в неё 75 кг воды из-под крана с температурой 16 °С, получить воду пригодную для купания ребёнка при 36 °0?

Answer 1

Ответ:

Из горячего озёра надо взять 125 л воды.

Объяснение:

Дано:

t₁ = 48 °С

m₂ = 75 кг

t₂ = 16 °С

t = 36 °C?

ρ = 1000 кг/м³ = 1 кг/дм³ = 1 кг/л

V₁ - ?

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Уравнение теплового баланса:

$Q_1+Q_2 = 0$

Q₁ - количество теплоты, отданное горячей водой:

$Q_1 = m_1c(t - t_1)$

Q₂ - количество теплоты, полученной водой из под крана:

$Q_2 = m_2c(t - t_2)$

Подставим:

$m_1c(t - t_1) + m_2c(t - t_2) = 0$

Сократим c и поменяем в первой скобке местами t и t₁:

$m_1(t_1 - t) = m_2(t - t_2)$

Масса горячей воды:

$m_1 = m_2 \dfrac{t - t_2}{t_1 - t}$

Чтобы получить объем в литрах надо массу разделить на плотность воды:

$V_1 = \dfrac{m_1}{\rho}$

$V_1 = \dfrac{m_2}{\rho} \cdot \dfrac{t - t_2}{t_1 - t}$

Расчет:

V₁ = (75 кг / 1 кг/л) · (36-16 °C) / (48-36 °C) = 75 л · 20 / 12 = 125 л

#SPJ1

Answer 2

Calculating the Amount of Water Needed

To calculate the amount of water needed to achieve a temperature of 36°C, we can use the principle of heat transfer. The heat gained or lost by an object can be calculated using the formula:

Q = mcΔT

Where: - Q is the heat gained or lost - m is the mass of the object - c is the specific heat capacity of the substance - ΔT is the change in temperature

In this case, we want to find the mass of water needed to achieve a temperature of 36°C when mixed with water at 48°C and water at 16°C.

Let's assume the specific heat capacity of water is 4.18 J/g°C.

First, let's calculate the heat gained by the water at 48°C when it cools down to 36°C:

Q1 = m1 * c * ΔT1

Where: - Q1 is the heat gained by the water at 48°C - m1 is the mass of the water at 48°C - c is the specific heat capacity of water - ΔT1 is the change in temperature from 48°C to 36°C

Next, let's calculate the heat gained by the water at 16°C when it heats up to 36°C:

Q2 = m2 * c * ΔT2

Where: - Q2 is the heat gained by the water at 16°C - m2 is the mass of the water at 16°C - c is the specific heat capacity of water - ΔT2 is the change in temperature from 16°C to 36°C

Since the total heat gained by the system is zero (assuming no heat is lost to the surroundings), we can set up the following equation:

Q1 + Q2 = 0

m1 * c * ΔT1 + m2 * c * ΔT2 = 0

Now, let's substitute the given values into the equation and solve for the unknown mass, m2:

m1 * c * (36 - 48) + m2 * c * (36 - 16) = 0

m1 * c * (-12) + m2 * c * 20 = 0

m1 * (-12) + m2 * 20 = 0

m2 = (m1 * 12) / 20

We know the mass of the water at 48°C is 75 kg. Let's substitute this value into the equation:

m2 = (75 * 12) / 20

m2 = 45 kg

Therefore, you would need to take 45 kg of water from the tap at 16°C and mix it with the water from the hot lake at 48°C to obtain water suitable for bathing a child at 36°C.

Please note that this calculation assumes no heat is lost to the surroundings and that the specific heat capacity of water remains constant over the temperature range.