(A) What is the required resistance of an immersion heater that will increase the temperature of 1.50 kg of water from 10.0°C to 50.0°C in 10.0 min while operating at 110 V?
(B) Estimate the cost of heating the water.

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Accepted Answer

This example allows us to link our new understanding of power in electricity with our experience with specific heat in thermodynamics (Chapter 20). An immersion heater is a resistor that is inserted into a container of water. As energy is delivered to the immersion heater, raising its temperature, energy leaves the surface of the resistor by heat, going into the water. When the immersion heater reaches a constant temperature, the rate of energy delivered to the resistance by electrical transmission is equal to the rate of energy delivered by heat to the water.

(A) To simplify the analysis, we ignore the initial period during which the temperature of the resistor increases, and also ignore any variation of resistance with temperature. Thus, we imagine a constant rate of energy transfer for the entire 10.0 min. Setting the rate of energy delivered to the resistor equal to the rate of energy entering the water by heat, we have

[latex] P =\frac{(\Delta V)^{2}}{R}=\frac{Q}{\Delta t} [/latex].

where Q represents an amount of energy transfer by heat into the water and we have used Equation 27.23 to express the electrical power. The amount of energy transfer by heat necessary to raise the temperature of the water is given by Equation 20.4, [latex] Q=m c \Delta T . ext { Thus, } [/latex]

[latex] P =I^{2} R=\frac{(\Delta V)^{2}}{R} [/latex] (27.23).

[latex] Q=m c \Delta T [/latex] (20.4).

[latex] \frac{(\Delta V)^{2}}{R}=\frac{m c \Delta T}{\Delta t} \quad \longrightarrow \quad R=\frac{(\Delta V)^{2} \Delta t}{m c \Delta T} [/latex].

Substituting the values given in the statement of the problem, we have

[latex] R=\frac{(110 V )^{2}(600 s )}{(1.50 kg )\left(4186 J / kg \cdot{ }^{\circ} C ight)\left(50.0^{\circ} C -10.0{ }^{\circ} C ight)} [/latex].

= 28.9 Ω.

(B) Because the energy transferred equals power multiplied by time interval, the amount of energy transferred is

[latex] P \Delta t=\frac{(\Delta V)^{2}}{R} \Delta t=\frac{(110 V )^{2}}{28.9 \Omega}(10.0 min )\left(\frac{1 h }{60.0 min } ight) [/latex].

= 69.8 Wh = 0.069 8 kWh.

If the energy is purchased at an estimated price of 10.0¢ per kilowatt-hour, the cost is

[latex] ext { Cost }=(0.0698 kWh )(\$ 0.100 / kWh )=\$ 0.00698 [/latex].

[latex] \approx 0.7 ot [/latex].

Question 27.8: (A) What is the required resistance of an immersion heater t...

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