#### Question

A family comes home from a long vacation with laundry to do and showers to take. The water heater has been turned off during the vacation. If the heater has a capacity of 50.0 gallons and a 4 800-W heating element, how much time is required to raise the temperature of the water from 20.0$^ { \circ } \mathrm { C }$ to 60.0$^ { \circ } \mathrm { C }$? Assume the heater is well insulated and no water is withdrawn from the tank during that time.

#### Solutions

Verified#### Step 1

1 of 5In this problem, water of volume $V = 50.0~\mathrm{gallons}$ is heated by a heating element of power $\mathscr{P} = 4800~\mathrm{W}$. The water is heated from $T_{1} = 20.0~\mathrm{^{\circ}C}$ to $T_{2} = 60.0~\mathrm{^{\circ}C}$. We calculate the time taken to heat the water.

#### Step 1

1 of 3For the heating process we'll have to provide heat $Q$, which will be given by using power $P$ for time $t$. Let's remember the definition of heating power:

$P:=\frac{Q}{t}.$

The time required to create heat $Q$ will therefore be

$t=\frac{Q}{P}.$

If heat $Q$ is added to mass $m$ of a material of specific heat $c$, the equation describing the situation will be

$Q=cm(T_2-T_1),$

where $T_1$ and $T_2$ are the initial and final temperatures, respectively.

In our case, we don't know the mass of water, but we know its volume. Remembering the definition of density, we have

$\rho =\frac{m}{V}.$

The mass will therefore be

$m=\rho V.$

Substituting, we have

$Q=c\rho V(T_2-T_1).$

Now we can substitute at the expression for the time, finding

$t=\frac{c\rho V(T_2-T_1)}{P}.$