Question

В калориметре при температуре 0°С находится 500 г воды и 300 г люда. Какая температура установиться в калориметре, если долить в него 100 г кипятка? 500 г кипятка?

Answer 1

1)-1градус а во 2-2грпдуса

Answer 2

Calculation of Final Temperature in the Calorimeter

To calculate the final temperature in the calorimeter, we can use the principle of conservation of energy. The heat gained by the water and lead in the calorimeter will be equal to the heat lost by the hot water that is poured into it.

Let's denote the initial temperature of the water and lead in the calorimeter as T1, the final temperature as T2, the mass of water as m1, the mass of lead as m2, and the mass of hot water poured into the calorimeter as m3.

According to the principle of conservation of energy, the heat gained by the water and lead can be calculated using the formula:

Q1 = m1 * c1 * (T2 - T1) + m2 * c2 * (T2 - T1)

where c1 and c2 are the specific heat capacities of water and lead, respectively.

The heat lost by the hot water can be calculated using the formula:

Q2 = m3 * c3 * (T2 - T_hot)

where c3 is the specific heat capacity of water and T_hot is the initial temperature of the hot water.

Since the heat gained and lost are equal, we can set up the equation:

m1 * c1 * (T2 - T1) + m2 * c2 * (T2 - T1) = m3 * c3 * (T2 - T_hot)

Now, let's substitute the given values into the equation:

- Initial temperature of the water and lead, T1 = 0°C - Mass of water, m1 = 500 g - Mass of lead, m2 = 300 g - Mass of hot water poured into the calorimeter, m3 = 100 g - Specific heat capacity of water, c1 = 4.18 J/g°C (source - Specific heat capacity of lead, c2 = 0.13 J/g°C (source - Specific heat capacity of water, c3 = 4.18 J/g°C (source - Initial temperature of the hot water, T_hot = 100°C

Substituting these values into the equation, we get:

500 * 4.18 * (T2 - 0) + 300 * 0.13 * (T2 - 0) = 100 * 4.18 * (T2 - 100)

Simplifying the equation, we have:

2090 * T2 + 39 * T2 = 418 * T2 - 41800

Combining like terms, we get:

2529 * T2 = 418 * T2 - 41800

Simplifying further, we have:

2111 * T2 = -41800

Dividing both sides by 2111, we get:

T2 = -41800 / 2111

Calculating this value, we find:

T2 ≈ -19.8°C

Therefore, the final temperature in the calorimeter, after pouring 100 g of boiling water, would be approximately -19.8°C.

Please note that the negative sign indicates that the final temperature is below freezing point, which is not physically possible. This suggests that there may be an error in the calculations or the given values.