Question 17.5: 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—air) from a chilly 10.0°C to 25.0°C? Take the molar specific heat at constant volume of air to be \frac{5}{2} R .
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.
(a) Compute the current and power output.
Apply Ohm’s law to get the current:
I=\frac{\Delta V}{R}=\frac{50.0 V }{8.00 \Omega}=6.25 A
Substitute into Equation 17.9 to find the power:
\mathscr{P} =I^{2} R=\frac{\Delta V^{2}}{R}
\mathscr{P} =I^{2} R=(6.25 A )^{2}(8.00 \Omega)=313 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.
\begin{aligned}Q &=\Delta U=n C_{v} \Delta T \\&=\left(2.50 \times 10^{3} mol \right)\left(\frac{5}{2} \cdot 8.31 J / mol \cdot K \right)(298 K -283 K ) \\&=7.79 \times 10^{5} J\end{aligned}
Divide the thermal energy by the power, to get the time:
t=\frac{Q}{ \mathscr{P} }=\frac{7.79 \times 10^{5} J }{313 W }=2.49 \times 10^{3} s
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 forty minutes. However, the calculation doesn’t take into account conduction losses. Recall that a 20-cm-thick concrete wall, as calculated in Chapter 11, permitted the loss of over two megajoules an hour by conduction!