Question

Чтобы получить молоко с температурой 65С, в горячее молоко массой 1,28 кг при температуре 95С хозяйка добавила холодное молоко при температуре 5С. Определите массу холодного молока, добавленного хозяйкой. Потерями энергии пренебрегите.

Answer 1

Нет спасибо но ты мою жизнь спсла

Answer 2

Answer in detail. To get milk with a temperature of 65°C, the hostess added cold milk at a temperature of 5°C to hot milk with a mass of 1.28 kg at a temperature of 95°C. Determine the mass of cold milk added by the hostess. Neglect energy losses.

To solve this problem, we can use the principle of conservation of energy, which states that the total amount of heat energy in the system before and after mixing the milks is the same. We can write the following equation:

$$Q_{hot} + Q_{cold} = Q_{mix}$$

where $Q_{hot}$ is the heat energy of the hot milk, $Q_{cold}$ is the heat energy of the cold milk, and $Q_{mix}$ is the heat energy of the mixture. We can also express the heat energy as the product of the mass, the specific heat capacity, and the change in temperature of the substance. For example, $Q_{hot} = m_{hot} \cdot c \cdot (T_{hot} - T_{mix})$, where $m_{hot}$ is the mass of the hot milk, $c$ is the specific heat capacity of milk, $T_{hot}$ is the initial temperature of the hot milk, and $T_{mix}$ is the final temperature of the mixture. Similarly, we can write $Q_{cold}$ and $Q_{mix}$ in terms of their respective masses and temperatures.

Substituting these expressions into the equation, we get:

$$m_{hot} \cdot c \cdot (T_{hot} - T_{mix}) + m_{cold} \cdot c \cdot (T_{cold} - T_{mix}) = m_{mix} \cdot c \cdot (T_{mix} - T_{mix})$$

Since the final temperature of the mixture is given as 65°C, we can simplify the equation as:

$$m_{hot} \cdot c \cdot (T_{hot} - 65) + m_{cold} \cdot c \cdot (5 - 65) = 0$$

We can also cancel out the specific heat capacity of milk, since it is the same for both milks, and rearrange the equation to solve for the mass of the cold milk:

$$m_{cold} = \frac{m_{hot} \cdot (T_{hot} - 65)}{65 - 5}$$

Now, we can plug in the given values of the mass and temperature of the hot milk:

$$m_{cold} = \frac{1.28 \cdot (95 - 65)}{65 - 5}$$

$$m_{cold} = \frac{1.28 \cdot 30}{60}$$

$$m_{cold} = 0.64$$

Therefore, the mass of the cold milk added by the hostess is 0.64 kg.