Question

A 1.19-kg aluminum pot contains 2.31 L of water. Both pot and water are initially at \(19.7^{\circ} \mathrm{C} .\) How much heat must flow into the pot and the water to bring their temperature up to \(95.0^{\circ} \mathrm{C}\) ? Assume that the effect of water evaporation during the heating process can be neglected and that the temperature remains uniform throughout the pot and the water.

Short answer

Answer: The total heat required is approximately 805,062.55 J.

Step-by-step solution

Finding the mass of water

Given the volume of water is 2.31 L, we need to convert it into mass. We know that the density of water is \(\rho = 1000 \frac{kg}{m^3}\) or \(1 \frac{kg}{L}\). Therefore, multiplying the volume by density, we can find the mass of water. $$m_{water} = \rho \times V = 1 \frac{kg}{L} \times 2.31 L = 2.31 kg$$

Determine the specific heat capacities

We need to find the specific heat capacities of aluminum and water. The specific heat capacity is a property of the material and given by: \(c_{Aluminum} = 903 \frac{J}{kg \cdot K}\) and \(c_{Water} = 4186 \frac{J}{kg \cdot K}\).

Calculate the change in temperature

We are given the initial temperature, \(T_{initial} = 19.7^{\circ} \mathrm{C}\), and the final temperature, \(T_{final} = 95.0^{\circ} \mathrm{C}\). Calculate the change in temperature for both the aluminum pot and water using the formula: \(\Delta T = T_{final} - T_{initial}\). $$\Delta T = 95.0^{\circ} \mathrm{C} - 19.7^{\circ} \mathrm{C} = 75.3^{\circ} \mathrm{C}$$

Calculate the heat required for the aluminum pot

Now we can use the equation for heat transfer to calculate the heat required to raise the temperature of the aluminum pot. $$Q_{Aluminum} = m_{Aluminum}c_{Aluminum}\Delta T = 1.19 kg \times 903 \frac{J}{kg \cdot K} \times 75.3 K$$ $$Q_{Aluminum} = 80594.57 J$$

Calculate the heat required for the water

Use the equation for heat transfer to calculate the heat required to raise the temperature of the water. $$Q_{Water} = m_{Water}c_{Water}\Delta T = 2.31 kg \times 4186 \frac{J}{kg \cdot K} \times 75.3 K$$ $$Q_{Water} = 724467.978 J$$

Calculate the total heat required

Finally, add the heat required for the aluminum pot and the water to get the total heat required. $$Q_{Total} = Q_{Aluminum} + Q_{Water} = 80594.57 J + 724467.978 J = 805062.548 J$$ The total heat required to bring the temperature of the aluminum pot and the water up to \(95.0^{\circ} \mathrm{C}\) is approximately \(805,062.55 J\).

Key concepts

Specific Heat Capacity

At the heart of understanding heat transfer in materials like the aluminum pot and water in our exercise is the concept of specific heat capacity. This property is crucial because it defines how much heat energy is needed to change the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin).

Imagine you're warming up different materials—each will require a different amount of energy to reach the same temperature based on their specific heat capacity. Metals like aluminum usually have a lower specific heat capacity, meaning they heat up and cool down more rapidly than water, a substance known for its high specific heat capacity.

This explains why in our exercise, even though both the aluminum pot and the water are heated by the same temperature change, water needs more heat to reach the target temperature. By quantifying this heat using specific heat capacity, we can calculate precise energy demands for a vast array of applications, from culinary endeavors to industrial processes.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In this context, we're particularly interested in the first law of thermodynamics, which posits that energy cannot be created or destroyed, only transformed.

When applying this to our exercise, the energy in the form of heat that we supply to the pot and water is not lost; rather, it is transformed into an increased internal energy of the substances, which we observe as a rise in temperature. The more heat we apply, the more we increase the internal energy, and the higher the temperature goes until we reach our desired temperature of 95.0°C.

By involving thermodynamics in heat calculations, we're not just looking at temperature change; we're delving into the fundamental energy exchanges and transformations that are taking place at the molecular level within the pot and water.

Heat Calculation

Heat calculation is essentially the application of specific heat capacities in thermodynamic equations to ascertain the amount of energy transfer required to achieve a certain temperature change. In our problem, we put heat calculation into practice by using the formula:
\[Q = mc\Delta T\]
where:

• \(Q\) is the heat energy transferred,
• \(m\) is the mass of the substance,
• \(c\) is the specific heat capacity, and
• \(\Delta T\) is the change in temperature.

Given the initial and final temperatures alongside the masses and specific heat capacities of our substances, we calculated the heat required for both the aluminum pot and the water separately, and then summed these quantities for the total heat required. Clear, step-by-step calculations guide us through complex problems and ensure that students precisely understand energy transfer in thermodynamic processes.