The battery in a laptop computer supplies 11.1 V and has a capacity of 56 W? h. In ordinary use, it is discharged after 4 hours. What is the average current drawn by the laptop, and what is the average rate of heat dissipation from it? You may assume that the temperature of the computer remains constant.

Question

What will be an ideal response?

Answer 1

If the total capacity of the battery is 56 Watt-hours, and it is drained in 4 hours, then the average discharge rate is 56/4 = 14 W, or 14 J/s. Remembering that power is current times voltage, the average current drawn by the laptop is 14 W/11.1 V = 1.26 A.

Taking the computer (not including the battery) as the system, and neglecting any change in internal energy of the computer, all of the electrical work supplied to the computer by the battery (at an average rate of 24.4 J/s) is ultimately converted to heat. That is, we can write the first law as



If U remains constant, thenQ? = ?? = ?14 W. When the computer is at steady state (not changing

temperature) 14 W must be dissipated into the surroundings as heat.

If we instead considered a system that includes the battery as well as the rest of the computer, then the work would be zero (the battery is now part of the system, rather than the surroundings). We could write the electrical potential energy of the battery (when fully charged) as

= 56 W-hr

Thus, the total energy of the system would include this as well as the internal energy of the computer and we could write the first law as

= Q + W

Taking W = 0, = 0, and = ?56 W-hr (when the battery is fully discharged, = 0), we have ?56 W-hr = Q

The total heat removal is 56 W-hr over a period of 4 hours, so the average rate of heat flow is

Q? = ?56/4 = ?14 W