posted by nanag .
in a certain system,the water used for storage is initially at 70 degree Celsius and it is required to provide an average of 7.5 kW to keep the house at an average of 18 degree Celsius for 3 days,how much water is needed.
Q = Pt = mc(T2-T1)
Q = heat/energy (in J)
P = Power (in W)
t = time (in s)
m = mass (in kg)
c = specific heat capacity (in J/kg-K) = 4.184 J/kg-K for water
T = temperature (in C or K)
we're solving for mass. substituting,
(7.5*1000)*(3*24*60*60) = m*4.184*(70-18)
1944000000 = 217.568m
m = 8935137.5 kg = 8935.14 tons of water