After a hot shower and dishwashing, there is "no hot water" left in the 50-gal (185 L) water heater. This suggests that the tank has emptied and refilled with water at roughly 10°C a) how much energy needed to reheat the water to 50°C?

Ok so, now a 185 L tank is empty, and has been filled completely with water at 10°C.

You want to know how much energy is need to raise the temperature of that 185 L of water to 50°C.

So therefore let the energy needed be represented by: #q# in KJ
#q=mcDeltat#
Where m is the mass, c is the specific heat of water, and #Deltat#is the change in temperature of water.

So:
#c=4.184 (KJ)/ (Kg°C)#
#Deltat=40°C#
#m= 185 cancel(L)* (1 Kg)/ (1 cancel(L))= 185 Kg#

Therefore:
#q= (185 cancel(Kg))((4.184 KJ)/ (cancel(Kg°C)))(40cancel(°C))#
#q= 30962 KJ#

Heat required #Q# is

#Q=msDelta T#
where #m# is mass of water, #s=4.186\ kJkg^-1"^@C^-1# is the specific heat of water, and #DeltaT# is the change in temperature.

(a) Mass of #185\ L# is approximately #=185\ kg#. Inserting values in above equation we get

#:. Q=185xx4.186xx40#
#=>Q= 30976.4\ kJ#

(b) How long would it take if the heater output is #9500\ W#?
We know that #1.00\ Js^-1# converts to #1\ W#
Assuming electrical heater transfers heat to the water at #100%# efficiency. Time #t# required is

#30976.4xx10^3=9500t#
#=>t=(30976.4xx10^3)/(9500xx60)=53.3\ mts#