Question

У ванну налили 10 л води при 5 С і 20 л води при 80 С. Перемішавши воду, виявили, що температура становить 55 С. Порівняти кількості теплоти, що віддала гаряча вода і прийняла холодна вода. Нагріваючи воду витратили 84 кДж теплоти. Який об'єм води необхідно взяти, щоб її нагріти на 20 С? Ложку масою 48 г за температури 29 С опустили в окріп масою 200 г. З якого матеріалу виготовлено ложку, якщо після встановлення теплової рівноваги температура води знизилася до 99 С ? (ПІДКАЗКА: Q перше = Q друге. Окремо записуємо кількість теплоти для ложки і кількість теплоти для води. Прирівнюємо їх і визначаємо матеріал ложки.)

Answer 1

Calculation of Heat Transferred in the Bathtub

To compare the amounts of heat transferred by the hot and cold water, we can use the principle of conservation of energy. The heat lost by the hot water should be equal to the heat gained by the cold water.

Let's calculate the heat transferred by the hot water first. We know that the initial temperature of the hot water is 80°C, and the final temperature after mixing is 55°C. The volume of hot water is 20 liters.

To calculate the heat transferred, we can use the formula:

Q = mcΔT

Where: - Q is the heat transferred - m is the mass of the water - c is the specific heat capacity of water - ΔT is the change in temperature

The specific heat capacity of water is approximately 4.18 J/g°C.

First, we need to convert the volume of water to mass using the density of water, which is approximately 1 g/mL.

Mass of hot water = Volume of hot water * Density of water

Mass of hot water = 20 liters * 1000 g/liter

Mass of hot water = 20,000 g

Now, we can calculate the heat transferred by the hot water:

Q_hot = m_hot * c * ΔT_hot

Q_hot = 20,000 g * 4.18 J/g°C * (55°C - 80°C)

Q_hot = -209,000 J

Next, let's calculate the heat transferred by the cold water. We know that the initial temperature of the cold water is 5°C, and the final temperature after mixing is 55°C. The volume of cold water is 10 liters.

Using the same formula as before, we can calculate the heat transferred by the cold water:

Q_cold = m_cold * c * ΔT_cold

Q_cold = 10,000 g * 4.18 J/g°C * (55°C - 5°C)

Q_cold = 2,090,000 J

Therefore, the heat transferred by the hot water is -209,000 J and the heat transferred by the cold water is 2,090,000 J.

Calculation of Volume of Water to Heat

To calculate the volume of water needed to heat it by 20°C, we can use the formula:

Q = mcΔT

Where: - Q is the heat transferred - m is the mass of the water - c is the specific heat capacity of water - ΔT is the change in temperature

We know that the heat transferred is 84 kJ (84,000 J), the change in temperature is 20°C, and we need to find the mass of water.

Let's rearrange the formula to solve for mass:

m = Q / (c * ΔT)

m = 84,000 J / (4.18 J/g°C * 20°C)

m ≈ 1,005 g

Therefore, approximately 1,005 grams (or 1.005 liters) of water need to be taken to heat it by 20°C.

Calculation of Spoon Material

To determine the material of the spoon, we can use the principle of thermal equilibrium. When the spoon and the water reach thermal equilibrium, the heat lost by the spoon should be equal to the heat gained by the water.

Let's calculate the heat transferred by the spoon first. We know that the initial temperature of the spoon is 29°C, and the final temperature after reaching thermal equilibrium is 99°C. The mass of the spoon is 48 g.

Using the same formula as before, we can calculate the heat transferred by the spoon:

Q_spoon = m_spoon * c_spoon * ΔT_spoon

Let's assume the specific heat capacity of the spoon is 0.385 J/g°C (typical value for stainless steel).

Q_spoon = 48 g * 0.385 J/g°C * (99°C - 29°C)

Q_spoon = 1,152 J

Now, let's compare the heat transferred by the spoon and the heat transferred by the water:

Q_spoon = Q_water

1,152 J = 2,090,000 J

Since the heat transferred by the spoon is significantly lower than the heat transferred by the water, we can conclude that the spoon is made of a material with a lower specific heat capacity than water.

Unfortunately, without additional information or further analysis, it is not possible to determine the exact material of the spoon.