Question

On a hot day, you take a six-pack of soda on a picnic, cooling it with ice. Each empty (aluminum) can weighs \(12.5 \mathrm{~g}\) and contains \(12.0\) oz of soda. The specific heat of aluminum is \(0.902 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\); take that of soda to be \(4.10 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) (a) How much heat must be absorbed from the six-pack to lower the temperature from \(25.0^{\circ}\) to \(5.0^{\circ} \mathrm{C}\) ? (b) How much ice must be melted to absorb this amount of heat? \(\left(\Delta H_{\mathrm{fus}}\right.\) of ise is given in Table 8.2.)

Short answer

Question: Calculate the total heat absorbed by a six-pack of soda cans to lower the temperature from \(25.0^{\circ}\) to \(5.0^{\circ} \mathrm{C}\) and determine the amount of ice needed to absorb that heat. Answer: (a) The total heat absorbed by the six-pack of soda cans is approximately \(168.9 \mathrm{~kJ}\). (b) \(506.5 \mathrm{~g}\) of ice must be melted to absorb the heat and lower the temperature of the six-pack from \(25.0^{\circ}\) to \(5.0^{\circ} \mathrm{C}\).

Step-by-step solution

Calculate the heat absorbed by each can of soda

To find the heat absorbed by each can of soda, we can use the equation for heat transfer: \(Q = mc\Delta T\) where \(Q\) is the heat transfer, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. We will calculate the heat transfer separately for aluminum and soda, and then combine them later to find the total heat transfer.

Calculate the heat absorbed by the aluminum cans

First, we will calculate the heat absorbed by the aluminum cans. We are given the mass of each aluminum can as \(12.5 \mathrm{~g}\) and the specific heat capacity of aluminum as \(0.902 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). The temperature change is the same for both the soda and cans, which is \((5 - 25)^{\circ} C = -20^{\circ} C\). Since we are cooling the cans, the temperature change will be negative. Now, we can plug the values and calculate the heat absorbed by one aluminum can: \(Q_{\text{aluminum}} = (12.5)(0.902)(-20) = -225.25 \mathrm{~J}\) Since there are 6 cans in the six-pack, the total heat absorbed by the aluminum cans is: \(Q_{\text{aluminum (total)}} = 6(-225.25) = -1351.5 \mathrm{~J}\)

Calculate the heat absorbed by the soda

Now, we will calculate the heat absorbed by the soda in a similar way. We are given the volume of each soda can as \(12.0 \text{ oz}\), which we need to convert to grams using the fact that 1 oz = 28.35 g, so we have \((12.0)(28.35) = 340.2\mathrm{~g}\) of soda in each can. The specific heat capacity of soda is given as \(4.10 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and the temperature change will be the same as before. Calculate the heat absorbed by one can of soda: \(Q_{\text{soda}} = (340.2)(4.10)(-20) = -27924.4 \mathrm{~J}\) Since there are 6 cans of soda in the six-pack, the total heat absorbed by the soda is: \(Q_{\text{soda (total)}} = 6(-27924.4) = -167546.4 \mathrm{~J}\)

Determine the total heat absorbed by the six-pack

Now, we will add the heat absorbed by the aluminum cans and the heat absorbed by the soda to find the total heat absorbed by the six-pack: \(Q_{\text{total}} = Q_{\text{aluminum (total)}} + Q_{\text{soda (total)}} = -1351.5 - 167546.4 = -168897.9 \mathrm{~J}\) (a) Therefore, the total heat absorbed by the six-pack to lower the temperature from \(25.0^{\circ}\) to \(5.0^{\circ} \mathrm{C}\) is approximately \(168.9 \mathrm{~kJ}\).

Determine the amount of ice needed to absorb the heat

To determine the amount of ice required to absorb the heat, we will use the following equation: \(m_{\text{ice}} = Q_{\text{total}} / \Delta H_{\text{fus}}\) We are given the heat of fusion for ice as \(333.5 \mathrm{~J} / \mathrm{g}\). Substitute the values and solve for the mass of ice: \(m_{\text{ice}} = -168897.9 \mathrm{~J} / 333.5 \mathrm{~J} / \mathrm{g} = 506.5 \mathrm{~g}\) (b) Therefore, \(506.5 \mathrm{~g}\) of ice must be melted to absorb the heat and lower the temperature of the six-pack from \(25.0^{\circ}\) to \(5.0^{\circ} \mathrm{C}\).

Key concepts

Specific Heat Capacity

Specific heat capacity is a key concept when dealing with heat transfer. It represents the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. The formula to calculate heat transfer using specific heat capacity is:

\( Q = mc\Delta T \)

where:

• \( Q \) is the amount of heat absorbed or released,
• \( m \) is the mass of the substance,
• \( c \) is the specific heat capacity,
• \( \Delta T \) is the temperature change.

This formula allows us to calculate the heat absorbed or lost by a particular material, like the aluminum cans in our soda picnic scenario.
Different materials require different amounts of heat to change their temperature, which is why each has a different specific heat capacity. For example, in this exercise, aluminum cans have a specific heat capacity of \(0.902 \mathrm{~J/g\cdot{ }^{\circ} \mathrm{C}}\), while soda has \(4.10 \mathrm{~J/g\cdot{ }^{\circ} \mathrm{C}}\). These values show soda needs more energy to change temperature compared to aluminum, meaning it's better at holding temperature—a reason it's used here to cool items.

Latent Heat

Latent heat refers to the energy absorbed or released by a substance during a phase change without a change in temperature. This concept plays a critical role when ice melts to absorb heat without changing its temperature.

The equation for latent heat exchange is:
\[ Q = m \Delta H_{\text{fus}} \]
where:

• \( Q \) is the heat absorbed or released,
• \( m \) is the mass of the substance,
• \( \Delta H_{\text{fus}} \) is the heat of fusion.

In the given exercise, the heat of fusion for ice is \(333.5 \mathrm{~J/g}\). This means that for each gram of ice that melts, it absorbs \(333.5 \mathrm{~J}\) of heat energy.
This absorption of heat is necessary to maintain the cold temperature of the soda cans effectively. Understanding latent heat helps in predicting how much ice is necessary to absorb a certain amount of heat and maintain desired temperatures.

Temperature Change

Temperature change is a straightforward concept but crucial in assessing energy transfer, especially in thermodynamics. It is the difference between the final and initial temperature of a substance—specifically the part of the equation \( \Delta T = T_{\text{final}} - T_{\text{initial}} \).

In the soda cooling exercise, the temperature change is from \(25.0^{\circ} \mathrm{C}\) to \(5.0^{\circ} \mathrm{C}\), resulting in a negative \(\Delta T\) of \(-20^{\circ}\mathrm{C}\). This shows the temperature decrease, indicating the need for cooling—a negative temperature change implies heat is being removed from the system.

Considering temperature changes helps understand how quickly or slowly a substance will react to added or removed heat. Larger masses or substances with higher specific heat capacities will see smaller temperature changes with the same heat added or removed. These principles combined are the foundation of effective heat management in everyday situations such as cooling your beverages with ice.