Answer:
Minimum electrical power required = 3.784 Watts
Minimum battery size needed = 3.03 Amp-hr
Explanation:
Temperature of the beverages, [tex]T_L = 36^0 F = 275.372 K[/tex]
Outside temperature, [tex]T_H = 100^0F = 310.928 K[/tex]
rate of insulation, [tex]Q = 100 Btu/h[/tex]
To get the minimum electrical power required, use the relation below:
[tex]\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[/tex]
V = 5 V
Power = IV
[tex]W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A[/tex]
If the cooler is supposed to work for 4 hours, t = 4 hours
[tex]I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr[/tex]
Minimum battery size needed = 3.03 Amp-hr
