Question

Два одинаковых теплонепроницаемых сосуда заполнены частично водой при комнатной температуре. Один сосуд заполнен водой наполовину, другой — на треть. В сосуды до краев доливают горячую воду из чайника. Найди температуру to воздуха в комнате, если известно, что после установления теплового равновесия температуры воды в сосудах оказались t1 48 °С и t2 56 °С. Теплоёмкостью сосудов и теплообменом с окружающей средой пренебречь.​

Answer 1

Ответ:

Первоначально, вода в сосудах имеет комнатную температуру t0.

(t + t0)/2 = 48; t + t0 = 96

(2t + t0)/3 = 56; 2t + t0 = 168

t0 = 24°

****

Доливали воду с температурой:

t = 72°

Answer 2

Problem Analysis

We have two identical, thermally insulated vessels partially filled with water at room temperature. One vessel is filled halfway, while the other is filled one-third of the way. Hot water is poured into both vessels until they are filled to the brim. We need to find the temperature of the air in the room after thermal equilibrium is reached. We can assume that the vessels have the same heat capacity and that there is no heat exchange with the surroundings.

Solution

To solve this problem, we can use the principle of conservation of energy. The total energy before and after pouring hot water into the vessels should remain the same.

Let's assume the initial temperature of the water in the vessels is T0, and the final temperature after pouring hot water is T.

The energy gained by the water in the first vessel (filled halfway) is given by: E1 = (1/2) * C * (T - T0) (Equation 1)

The energy gained by the water in the second vessel (filled one-third of the way) is given by: E2 = (2/3) * C * (T - T0) (Equation 2)

Since the vessels are thermally insulated, the total energy gained by the water in both vessels should be equal to the energy lost by the hot water poured into the vessels.

Let's assume the mass of hot water poured into each vessel is m, and its initial temperature is Th. The energy lost by the hot water is given by: E_hot_water = 2 * m * (Th - T) (Equation 3)

Setting Equations 1 and 2 equal to Equation 3, we can solve for T.

Calculation

Let's substitute the given values into the equations and solve for T.

Given: T0 = room temperature T1 = 48 °C (temperature of water in the vessel filled halfway) T2 = 56 °C (temperature of water in the vessel filled one-third of the way)

We can rewrite Equations 1 and 2 as: E1 = (1/2) * C * (T - T0) (Equation 1') E2 = (1/3) * C * (T - T0) (Equation 2')

Substituting the given values into Equations 1' and 2', we get: (1/2) * C * (T - T0) = (1/3) * C * (T - T0)

Simplifying the equation, we have: (T - T0) / 2 = (T - T0) / 3

Multiplying both sides by 6 to eliminate the fractions, we get: 3(T - T0) = 2(T - T0)

Simplifying further, we have: 3T - 3T0 = 2T - 2T0

Combining like terms, we get: T = T0

Therefore, the temperature of the air in the room after thermal equilibrium is reached is equal to the initial temperature of the water in the vessels (T0).

Answer

The temperature of the air in the room, after thermal equilibrium is reached, is equal to the initial temperature of the water in the vessels, which is the room temperature.