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

Minimum electrical power required = 3.784 Watts

Minimum battery size needed = 3.03 Amp-hr

Step-by-step 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:

`(T_L)/(T_H - T_L) = (Q)/(W) \\W = (Q(T_H - T_L))/(T_L)\\W = (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) = (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