The capacity of batteries such as the typical AA alkaline battery is expressed in units of milliamp-hours (mAh). An AA alkaline battery yields a nominal capacity of \(2850 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of \(\mathrm{mAh} ?\) (b) The starting voltage of a fresh alkaline battery is \(1.55 \mathrm{~V}\). The voltage decreases during discharge and is \(0.80 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

mA (milliampere) is a measure of electric current, and h (hour) is a measure of time. Together, the unit mAh refers to the amount of electric charge the battery can deliver over a specified period ("hour" in this case). It essentially represents the battery life or how long the battery can provide power before it needs to be replaced or recharged. The greater the mAh value, the longer the battery can supply power.

Since the voltage declines linearly during discharge, we can represent the voltage versus charge curve as a trapezoid. To compute the maximum electrical work the battery can perform, we will find the area under this curve. The area of a trapezoid can be calculated as follows: \[A = \frac{1}{2}(Base_1 + Base_2) \times Height\] In this case, \(Base_1\) is the starting voltage \(1.55V\), \(Base_2\) is the ending voltage \(0.80V\), and the \(Height\) is the entire capacity in hours, which is \(\frac{2850 mAh}{1000mA/h}\) = \(2.85 h\). Now, we calculate the area: \[A = \frac{1}{2}(1.55V + 0.80V) \times 2.85 h\] \[A = \frac{1}{2}(2.35V) \times 2.85 h\] \[A = 3.34875 Vh\] Since work done is equal to charge times voltage, we can convert the obtained area value from "Vh" to Joules (the unit of energy) using the formula: \[Work = A \times Q\] where \(Q\) is the total charge in Coulombs. We are given the battery's capacity as \(2850 mAh = 2.85 Ah\). One ampere-hour is equal to 3600 Coulombs, so: \[Q = 2.85 Ah \times 3600 \frac{C}{Ah} = 10260 C\] Now, we can find the work: \[Work = 3.34875 Vh \times 10260 C\] \[Work \approx 34374.225 J\] So, the total maximum electrical work the battery could perform during discharge is approximately \(34374.225 Joules\).