Question

.. CP In the circuit in Fig. E26.19, a $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{\Omega }$resistor is inside 100gof pure water that is surrounded by insulating styrofoam.

If the water is initially at, how long will it take for its temperature to rise to$\mathbf{58}\mathbf{°}\mathbf{C}$?

Short answer

time required is t = 1006 s

Step-by-step solution

Step 1:About resitance in parrallel

theequivalentresistorofthecircuit
For the three branches in the circuit, each branch has two resistors in series; 80, the equivalent resistor of each branch is the
sum of the two resistors as shown by step (1) in the ?gure below.
Now, we have three resistors in parallel; Using equation 262, the equivalent resistor is given as followg
$\frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{20}+\frac{1}{20}+\frac{1}{10}$

$=\frac{1}{5\mathrm{\Omega }}$

${\mathrm{R}}_{\mathrm{eq},1}=5\mathrm{\Omega }$

have three resistors in series, thus the equivalent resistor of the circuit is
${\mathrm{R}}_{\mathrm{eq}}=20+5+5=30\mathrm{\Omega }$
Using Ohm‘s law, the current of the circuit is;
$$\begin{gathered} \mathrm{l}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathrm{eq}}} \\ =\frac{30\mathrm{V}}{30\mathrm{\Omega }} \\ =1\mathrm{A} \end{gathered}$$

Sincetheresistor20Qisinseries,thesamecurrentis?owingthroughit.
From equation (25.18), the pOWer dissipated in a resistor R with current I flowing through is given by:
$\mathrm{P}={\mathrm{I}}^2\mathrm{R}$
Now, we plug our values for R20 Q and I, so We get the pOWer dissipated in the resistor 20 Q:

r 20 Q:

$$\begin{gathered} \mathrm{P}=1^2\times 20 \\ =20\mathrm{J}/\mathrm{s} \end{gathered}$$
This is the rate of energy change from electic energy to thermal energy

Determine the time required

“’0know.fromequation17.13,theenergyrequiredtocauseatemperature
change AT of a substance of speci?c heat. I: and mass m is given by:
$\mathrm{E}=\mathrm{mc}∆\mathrm{T}$
“’0 know, from equation 17.131 the energy required to cause a temperature
change AT of a substance of speci?c heat. c and mass m is given by:

For the water surrounding the resistor ‘20 52, we have

$\mathrm{m}=0.100\mathrm{kg},\mathrm{c}=4190\mathrm{J}/\mathrm{kg},∆\mathrm{T}=48$
’I‘lms, the energy e’lsferred from the registor 1.0 the water is$\mathrm{F}=20112\mathrm{J}$
Now, we lmve Llu‘e power and the total energy required L0 raise the temperature
of the water, the time rm‘luired for this process is:
$$\begin{gathered} \mathrm{t}=\frac{\mathrm{E}}{\mathrm{P}}=\frac{20112}{20} \\ =1005.6\mathrm{s} \end{gathered}$$

therefore time required is t = 1006 s