# Question 3629f

See example below:

#### Explanation:

Supose you want to heat 2 liters of water using an electric heater.
The water has to be heated from ${20}^{o} C$ to ${80}^{o} C$ and the heater has a resistance of 30 ohms to be connected to a power supply of 120Volts.
The heater is thermal insulated from the environment.
How much time wil be taken to do the job?

First, we have to calculate the work, in calories:

$\tau$ = m.c. $\Delta \theta$

where:

$\tau$ = work in calories
m = mass in (g)
c = specific heat of water = 1
$\Delta \theta$ = temperature difference (theta_2-theta_1)#
$\Delta \theta$ = $80 - 20 = {60}^{o} C$

As the water specific mass=1, 2 liters of water have 2000g
(m=2000g).

$\tau = 2000 \times 1 \times 60$ = 120000 calories

The electrical power of the heater is:

$P = {V}^{2} / R = {120}^{2} / 30 = 480$ Watts

The electrical work in Joules is equal to:

$W = P . t$

where:

W = Electrical Work, in Joules
P = Electrical Power, in Watts
t = time, in seconds

To convert calories in Joules, we use the equation:

$\tau = 0.239 \times P \times t$

$120000 = 0.239 \times 480 \times t$

$t = \frac{120000}{0.239 \times 480}$

$t \cong 1046 s \cong 17 \min 26 s$