**Heating a Water Bath**

**Solutions Below:**

Part A

Use 4190 J/kg°C as the heat capacity of water, and express your answer in seconds using three significant figures.

R1 and R2 (see picture for labels) are in series. R3 and R4 are in another series, and R5 and R6 are in a 3rd series. These 3 series are in parallel with each other, and the whole parallel circuit is in series with RWater and R7. Find the 3 smaller series first:

R_{12} = R1 + R2

R_{12} = 10.0 + 20.0

R_{12} 20.0

Likewise, R_{34} = 20.0 and R_{56} = 10.0. The parallel circuit is given by:

R_{Parallel} = 1 / ((1 / R_{12}) + (1 / R_{34}) + (1 / R_{56}))

R_{Parallel} = 1 / ((1 / 20.0) + (1 / 20.0) + (1 / 10.0))

R_{Parallel} 5.0

And the series between RWater, R_{Parallel} and R7:

RTotal = RWater + R_{Parallel} + R7

RTotal = 20.0 + 5.0 + 5.0

RTotal = 30.0 Ω

Next, find the current:

V = IR

30.0 = I * 30.0

I = 1.00 A

Next, find the voltage drop across the resistor in the water:

V = IR

V = 1.00 * 20.0

V = 20.0

Now find the power dissipated in the water. There are a couple formulas you can use but the most convenient is W = V^{2} / R. You want the voltage dissipated in the water and the resistance of the resistor in the water:

W = V^{2} / R

W = (20.0)^{2} / 20.0

W = 20.0

One watt is one joule per second. So find the total number of joules needed to raise the temperature. The temperature change is 57.3 – 11.7 = 45.6. There are 0.103 kg of water (103 g / 1000). And the specific heat to use is 4190 J / kg°C:

J = mcΔT

J = 0.103 * 4190 * 45.6

J = 19680

Since there are 20.0 watts of power, the time needed (in seconds) is 19680 / 20 = 984 s

984 s

Great explanation! Very easy to follow. Thanks!

Thanks! Glad to help!