Question 17.6: The Power Converted by an Electric Heater GOAL Calculate an ...
The Power Converted by an Electric Heater
GOAL Calculate an electrical power output and link to its effect on the environment through the first law of thermodynamics.
PROBLEM An electric heater is operated by applying a potential difference of 50.0 V to a Nichrome wire of total resistance 8.00 Ω. (a) Find the current carried by the wire and the power rating of the heater. (b) Using this heater, how long would it take to heat 2.50 × 10³ moles of diatomic gas (e.g., a mixture of oxygen and nitrogen, or air) from a chilly 10.0°C to 25.0°C? Take the molar specific heat at constant volume of air to be \textstyle{\frac{5}{2}} R. (c) How many kilowatt-hours of electricity are used during the time calculated in part (b) and at what cost, at $0.12 per kilowatt-hour?
STRATEGY For part (a), find the current with Ohm’s law and substitute into the expression for power. Part (b) is an isovolumetric process, so the thermal energy provided by the heater all goes into the change in internal energy, ΔU. Calculate this quantity using the first law of thermodynamics and divide by the power to get the time. Finding the number of kilowatt-hours used requires a simple unit conversion technique. Multiplying by the cost per kilowatt yields the total cost of operating the heater for the given time.
(a) Compute the current and power output.
Apply Ohm’s law to get the current:
I={\frac{\Delta V}{R}}={\frac{50.0\,\mathrm{V}}{8.00\,{\Omega}}}=\ \ {6.25\,}A
Substitute into Equation 17.9 to find the power:
P=I^{2}R=(6.25\,\mathrm{A})^{2}(8.00\,\Omega)\,=\,313\,\mathrm{W}
(b) How long does it take to heat the gas?
Calculate the thermal energy transfer from the first law.
Note that W = 0 because the volume doesn’t change.
Q=\Delta U=\ n C_{v}\Delta T
=(\ 2.50\times10^{3}\mathrm{~mol})({\frac{5}{2}}\cdot8.31\,\mathrm{~J}/{\mathrm{mol}}\cdot\mathrm{K})(298\mathrm{~K}-\ 283\mathrm{~K})
=7.79\times10^{5}~J
Divide the thermal energy by the power to get the time:
t={\frac{Q}{P}}={\frac{7.79~\times~10^{5}\mathrm{J}}{313\,\mathrm{W}}}=\ {2.49\times10^{3}\,\mathrm{s}}
(c) Calculate the kilowatt-hours of electricity used and the cost.
Convert the energy to kilowatt-hours, noting that 1 J = 1 W · s:
U=(7.79\times10^{5}\,\mathrm{W}\cdot{\mathrm{s}})\biggl({\frac{1.00\,\,{\mathrm{kW}}}{1.00~\times~10^{3}\,\,{\mathrm{W}}}}\biggr)\biggl({\frac{1.00\,\,{\mathrm{h}}}{3.60~\times~10^{3}\,{\mathrm{s}}}}\biggr)={{0.216\,\,{\mathrm{kWh}}}}
Multiply by $0.12/kWh to obtain the total cost of operation:
Cost = (0.216 kWh)($0.12/kWh) =$0.026
REMARKS The number of moles of gas given here is approximately what would be found in a bedroom. Warming the air with this space heater requires only about 40 minutes. The calculation, however, doesn’t take into account conduction losses. Recall that a 20-cm-thick concrete wall, as calculated in Chapter 11, permitted the loss of more than 2 megajoules an hour by conduction!