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.. CP In the circuit in Fig. E26.19, a 20.0Ωresistor 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 to58°C?

Short Answer

Expert verified

time required is t = 1006 s

Step by step solution

01

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
1Req=120+120+110

=15Ω

Req,1=5Ω

have three resistors in series, thus the equivalent resistor of the circuit is
Req=20+5+5=30Ω
Using Ohm‘s law, the current of the circuit is;
l=VReq=30V30Ω=1A

Sincetheresistor20Qisinseries,thesamecurrentis?owingthroughit.
From equation (25.18), the pOWer dissipated in a resistor R with current I flowing through is given by:
P=I2R
Now, we plug our values for R20 Q and I, so We get the pOWer dissipated in the resistor 20 Q:

r 20 Q:

P=12×20=20J/s
This is the rate of energy change from electic energy to thermal energy

02

Determine the time required

“’0know.fromequation17.13,theenergyrequiredtocauseatemperature
change AT of a substance of speci?c heat. I: and mass m is given by:
E=mcT
“’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

m=0.100kg,c=4190J/kg,T=48
’I‘lms, the energy e’lsferred from the registor 1.0 the water isF=20112J
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:
t=EP=2011220=1005.6s

therefore time required is t = 1006 s

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