tem 4 Part A If the water is initially at temperature 12.0 ° C, how long will it

Question

tem 4 Part A If the water is initially at temperature 12.0 ° C, how long will it take for its temperature to rise to 58.5 ° C? Use 4190 J/kg.C° as the heat capacity of water, and express your answer in seconds using three significant figures. Constants In the circuit in the figure, (Figure 1)a 20-ohm resistor sits inside 108 g of pure water that is surrounded by insulating Styrofoam View Available Hint(s) Submit Provide Feedback Figure of 1 10.012 10.012 20.0 ? 10.0 ? 10.0 ? Water 5.0 ? 5.0 ? 0.0 V 5.0?

Explanation / Answer

Energy required to increase the temperature of water is given by:

Q = m*C*dT

Using given values:

Q = 0.108 kg*4190 J/kg-C*(58.5 - 12) C

Q = 21042.18 J

Now in the given Circuit

Remember:
For series combination
Req = R1 + R2 + R3 +...............
for parallel combination
1/Req = 1/R1 + 1/R2 + 1/R3 + ............
for 2 resistors in parallel it will be
Req = R1*R2/(R1+R2)
Using this Information:

10 and 10 ohm are in series, So,

Rs1 = 10 + 10 = 20 ohm

5 and 5 ohm are in series, So,

Rs2 = 5 + 5 = 10 ohm

Now Rs1, Rs1, and Rs2 are in parallel, SO

1/Rp = 1/10 + 1/20 + 1/20 = 4/20

Rp = 5 ohm

Now Rp, 20 ohm (left) and 5 ohm (bottom right) are in series, So

Req = 5 + 20 + 5

Req = 30 ohm

ieq = V/Req = 30/30 = 1 Amp

Remember in resistors parallel combination voltage distribution in each part will be same and in series combination current distribution in each resistor will be same.

Current in 20 ohm (left) = ieq = 1 Amp

Power in 20 ohm (left) = i^2*R

P = 1^2*20 = 20 W

Now we know that

P = E/t

t = E/p = 21042.18/20

t = 1052.11 sec

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