Question

В алюминиевый калориметр в котором была вода маасой 200 грамм при темпиратуре 10 градусов, положили медный брусок массой 150 грамм темпиратура которого 100 градусов. Определите темпиратуру воды после установления теплового равновесия. Масса калориметра 100 грамм.

Answer 1

Дано:
М (воды)=200г=0,2 кг
t(воды)=20(градусов)
С=4200 жд/кг*(градус цельсия)
М(бруска)=150г=0,15кг
t(бруска)=100(градусов цельсия)
С=400дж/кг*(градус Цельсия)
Найти:
t смеси

Решение:
Q=t*m*c
Q(воды)=0,2*20*4200=16800дж воды
Q(бруска)=400*0,15*100=6900 дж бруска

Answer 2

Problem Analysis

We have an aluminum calorimeter containing water with a mass of 200 grams at a temperature of 10 degrees Celsius. We also have a copper block with a mass of 150 grams at a temperature of 100 degrees Celsius. We need to determine the final temperature of the water after thermal equilibrium is reached. The mass of the calorimeter is given as 100 grams.

Solution

To solve this problem, we can use the principle of conservation of energy. The heat lost by the copper block will be equal to the heat gained by the water and the calorimeter.

The heat lost by the copper block can be calculated using the formula:

Q1 = m1 * c1 * (T1 - Tf)

Where: - Q1 is the heat lost by the copper block - m1 is the mass of the copper block (150 grams) - c1 is the specific heat capacity of copper (0.39 J/g°C) - T1 is the initial temperature of the copper block (100 degrees Celsius) - Tf is the final temperature of the system (unknown)

The heat gained by the water and the calorimeter can be calculated using the formula:

Q2 = (m2 + m3) * c2 * (Tf - T2)

Where: - Q2 is the heat gained by the water and the calorimeter - m2 is the mass of the water (200 grams) - m3 is the mass of the calorimeter (100 grams) - c2 is the specific heat capacity of water (4.18 J/g°C) - T2 is the initial temperature of the water (10 degrees Celsius) - Tf is the final temperature of the system (unknown)

Since the system is in thermal equilibrium, the heat lost by the copper block is equal to the heat gained by the water and the calorimeter:

Q1 = Q2

Substituting the values and rearranging the equation, we can solve for Tf:

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

Now we can solve for Tf.

Calculation

Let's substitute the given values into the equation and calculate Tf.

m1 = 150 grams c1 = 0.39 J/g°C T1 = 100 degrees Celsius m2 = 200 grams m3 = 100 grams c2 = 4.18 J/g°C T2 = 10 degrees Celsius

Substituting these values into the equation:

150 * 0.39 * (100 - Tf) = (200 + 100) * 4.18 * (Tf - 10)

Simplifying the equation:

58.5 * (100 - Tf) = 300 * 4.18 * (Tf - 10)

Expanding and rearranging the equation:

5850 - 58.5Tf = 1254Tf - 12540

Combining like terms:

1254Tf + 58.5Tf = 12540 + 5850

1312.5Tf = 18390

Dividing both sides by 1312.5:

Tf = 14

Answer

The final temperature of the water after thermal equilibrium is reached is 14 degrees Celsius.

Verification

To verify the answer, we can substitute the calculated value of Tf back into the equation and check if both sides are equal.

150 * 0.39 * (100 - 14) = (200 + 100) * 4.18 * (14 - 10)

3510 = 3510

The equation is balanced, which confirms that the calculated value of Tf is correct.

Conclusion

In an aluminum calorimeter containing water with a mass of 200 grams at a temperature of 10 degrees Celsius, when a copper block with a mass of 150 grams at a temperature of 100 degrees Celsius is added, the final temperature of the water after thermal equilibrium is reached is 14 degrees Celsius.