A system consists of a 0.130-kg chunk of ice floating in 1.12 kg of water, all at 0 ∘C. Part A How much thermal energy must be added to the system to convert it to 1.25 kg of water at 15.0 ∘C?
Heatfusion*.130 + 1.25*Cwater*(15-0)=?
To solve this problem, we can break it down into several steps:
Step 1: Calculate the energy required to heat the ice to its melting point (0 degrees Celsius):
The specific heat capacity of ice is 2.09 J/g°C.
The initial temperature of the ice is 0°C.
The final temperature of the ice is 0°C.
The mass of the ice is 0.130 kg.
Using the equation:
Q = m * C * ∆T,
where Q is the thermal energy, m is the mass, C is the specific heat capacity, and ∆T is the change in temperature.
Q = 0.130 kg * 2.09 J/g°C * (0°C-0°C)
Q = 0 J
Therefore, no energy is required to heat the ice to its melting point since the ice is already at 0°C.
Step 2: Calculate the energy required to melt the ice:
The heat of fusion for water is 334 J/g.
The mass of the ice is 0.130 kg.
Using the equation:
Q = m * H,
where Q is the thermal energy, m is the mass, and H is the heat of fusion.
Q = 0.130 kg * 334 J/g
Q = 43.42 J
Therefore, 43.42 J of thermal energy is required to melt the ice.
Step 3: Calculate the energy required to heat the melted water from 0°C to 15°C:
The specific heat capacity of water is 4.18 J/g°C.
The initial temperature of the water is 0°C.
The final temperature of the water is 15°C.
The mass of the water is 1.12 kg.
Using the equation:
Q = m * C * ∆T,
where Q is the thermal energy, m is the mass, C is the specific heat capacity, and ∆T is the change in temperature.
Q = 1.12 kg * 4.18 J/g°C * (15°C - 0°C)
Q = 88.428 J
Therefore, 88.428 J of thermal energy is required to heat the water from 0°C to 15°C.
Step 4: Calculate the energy required to heat the water from 0°C to 15°C:
The specific heat capacity of water is 4.18 J/g°C.
The initial temperature of the water is 0°C.
The final temperature of the water is 15°C.
The mass of the water is 1.25 kg.
Using the equation:
Q = m * C * ∆T,
where Q is the thermal energy, m is the mass, C is the specific heat capacity, and ∆T is the change in temperature.
Q = 1.25 kg * 4.18 J/g°C * (15°C - 0°C)
Q = 97.875 J
Therefore, 97.875 J of thermal energy is required to heat the water from 0°C to 15°C.
Step 5: Sum up the thermal energies from the previous steps to calculate the total energy required to convert the system to 1.25 kg of water at 15.0°C:
Total energy = Energy to heat ice to melting point + Energy to melt ice + Energy to heat melted water to 15°C + Energy to heat 1.25 kg of water to 15°C.
Total energy = 0 J + 43.42 J + 88.428 J + 97.875 J
Total energy = 229.723 J
Therefore, the total thermal energy that must be added to the system to convert it to 1.25 kg of water at 15.0°C is approximately 229.723 J.
To calculate the thermal energy required to convert the system from the initial state to the final state, you need to consider the following steps:
Step 1: Calculate the heat energy required to warm the ice from 0 °C to its melting point.
The specific heat capacity of ice is c_ice = 2.09 J/g °C, and the mass of the ice is 0.130 kg. The initial temperature of the ice is 0 °C, and the final temperature is the melting point of ice, which is 0 °C.
To calculate the heat energy required to warm the ice, use the formula:
Q_1 = (mass of ice) × (specific heat capacity of ice) × (change in temperature)
Q_1 = (0.130 kg) × (2.09 J/g °C) × (0 °C)
Step 2: Calculate the heat energy required to melt the ice.
The heat of fusion for water is L_fusion = 334 J/g. Since the ice is at its melting point (0 °C), we can use the following formula to calculate the heat energy required to melt the ice:
Q_2 = (mass of ice) × (heat of fusion)
Q_2 = (0.130 kg) × (334 J/g)
Step 3: Calculate the heat energy required to warm the resulting water from 0 °C to 15.0 °C.
The specific heat capacity of water is c_water = 4.18 J/g °C, and the mass of the water is (initial mass of water) + (mass of melted ice). The initial temperature is 0 °C, and the final temperature is 15.0 °C.
To calculate the heat energy required to warm the water, use the formula:
Q_3 = (mass of water) × (specific heat capacity of water) × (change in temperature)
Q_3 = [(1.12 kg) + (0.130 kg)] × (4.18 J/g °C) × (15.0 °C)
Step 4: Sum up the heat energies calculated in the previous steps to obtain the total thermal energy required.
Total thermal energy = Q_1 + Q_2 + Q_3
Once you perform these calculations, you will have the answer to Part A of the question, which asks for the amount of thermal energy needed to convert the system to 1.25 kg of water at 15.0 °C.