Question

A household in Albuquerque uses 100 gallons of hot water per day (1 gallon of water has a mass of 3.78 kg), and the water temperature must be raised from 10°C to 50°C. (a) Find the average power needed to heat the water. (b) Assume the house has solar collectors with 55% average efficiency at converting sunlight to thermal energy when tilted at Albuquerque’s latitude. Use data from Table 9.1 to nd the collector area needed to supply this power in June.

Answer 1

According to the concpt of the thermal propertices of mater

Given that

mass m=100*3.78=378 kg

time t=86400 sec

temperature difference =(T2-T1)=(273+50)-(273+10)=40 k

(T2-T1)=40 K

specific heat of the water =4200 J/kg.K

now we find the average power needed to heat work

Power =ms(T2-T1)/t

=378*4200*40/86400

=735 W

now we find the output power

55/100=Pout/p

Pout=55*735/100=404.25 W

now we find the area

Pout=A*sigma*(T^4

404.25=A*5.67*10^-8*40^4

Area A=2785.m^2