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.)

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.

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: \(Q_{\text{aluminum (total)}} = 6(-225.25) = -1351.5 \mathrm{~J}\)

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: \(Q_{\text{soda (total)}} = 6(-27924.4) = -167546.4 \mathrm{~J}\)

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 \(168.9 \mathrm{~kJ}\).

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 \(5.0^{\circ} \mathrm{C}\).