Question

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?

Answer 1

`q= 30962 KJ`

Explanation:

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`

Answer 2

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`