Question

1. A team of students have designed a battery-powered cooler, which promises to keep beverages at a high-drinkability temperature of 36°F while outside is 100°F. If the insulation leaks heat at a rate of 100 Btu/h, calculate the minimum electrical power required, in Watts. If we use USB-charged 5 V batteries, what is the minimum battery size needed, in Amp-hours, if the cooler is supposed to work for 4 hours?

Answer 1

Answer:

Minimum electrical power required = 3.784 Watts

Minimum battery size needed = 3.03 Amp-hr

Explanation:

Temperature of the beverages, $T_L = 36^0 F = 275.372 K$

Outside temperature, $T_H = 100^0F = 310.928 K$

rate of insulation, $Q = 100 Btu/h$

To get the minimum electrical power required, use the relation below:

$\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt$

V = 5 V

Power = IV

$W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A$

If the cooler is supposed to work for 4 hours, t = 4 hours

$I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr$

Minimum battery size needed = 3.03 Amp-hr