In a passive solar house, the sun heats water stored in barrels to a temperature of 38°C.The stored energy is

Question

then used to heat the house on cloudy days. Suppose that 2.4×10^8 J of heat are needed to maintain the inside of the house at 21°C. how many barrels ( 1 BARREL= 0.16 M^3) OF WATER are needed?

What does the E stand for

edwards what is E

Answer 1

We have to determine the heat capacity of one barrel and divide the total heat required [ 2.4E+8J ]over that number.
The heat capacity per barrel is
Q(barrel)= m Cp (T2-T1)
Q(barrel)= pV Cp (T2-T1) since Cp=4181.3 kJ/kg K

Q(barrel)= 1E+3 x 0.16 4181.3 (38 - 21)=
Q(barrel)= 11 E+6 J

Number of barrels= 2.4E+8/11.4 E+6 = 21 barrels

Answer 2

Available energy in joules = heat capacity * mass * temperature delta
Heat capacity = 4.2J/g per degrees Celsius.
Mass = 1m^3 of water weights 1000kg so 0.16m^3 weights 160000 grams.
Temperature delta = 38-21=17
==11megajoules of energy is stored in a barrel

So it would be like running a 1kw space heater for 2.75 hours (a joule is a watt second)
I can't say how many btu's you need to heat your house, use this calculator. http://hearth.com/calc/btucalc.html

3.4watts equals a btu/hour.