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My dad told me to turn off the lights when I'm not using them because it will keep the house cooler. I don't think he is correct that a light can heat up the room. How much of a difference in the temperature would it make? Assume a 60 Watt light bulb running for one hour in a room with approx. 3000ft^3/85m^3 of air. Also assume the air is dry and in constant thermal equilibrium at 1 atm. I think the specific heat capacity of air is about 1 J/g*K, but I am not confident.

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  • $\begingroup$ Looks like you have all figures you need. Why wouldn't you plug them into the formula and do the math? $\endgroup$ – Ivan Neretin Aug 28 '15 at 11:15
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During a period of $\Delta t = 1\ \mathrm h =3600\ \mathrm s$, the light bulb converts electric power of $P = 60\ \mathrm W$ to light and heat. Assuming that none of this energy escapes from the room, the entire power is eventually converted to heat:

$$\begin{align} Q &= P \cdot \Delta t \\ &= 60\ \mathrm W \times 3600\ \mathrm s \\ &= 216\ \mathrm{kJ} \end{align}$$

According to REFPROP (NIST Standard Reference Database 23, Version 9.0), the relevant parameter values of air at the assumed initial temperature of $T_0 = 20\ \mathrm{^\circ C}$ and pressure $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$ are a density of $\rho_0 = 1.2046\ \mathrm{kg/m^3}$ and a specific heat capacity at constant pressure of $c_{p,0} = 1.0061\ \mathrm{kJ/(kg\ K)}$.

The initial volume is given as $V_0 = 85\ \mathrm{m^3}$. The corresponding mass is

$$\begin{align} m &= V_0 \cdot \rho_0 \\ &= 85\ \mathrm{m^3} \times 1.2046\ \mathrm{kg/m^3} \\ &= 102.391\ \mathrm{kg} \end{align}$$

Ignoring the heat capacity of the walls and inventory, the temperature of the air is approximately increased by

$$\begin{align} \Delta T &= \frac QC \\ &= \frac Q{m \cdot c_{p,0}} \\ &= \frac{216\ \mathrm{kJ}}{102.391\ \mathrm{kg} \times 1.0061\ \mathrm{kJ/(kg\ K)}} \\ &= 2.1\ \mathrm{K} \end{align}$$

By way of comparison, a better result may be obtained using the enthalpy of air, since the heat capacity of air is not constant. In this case, however, the difference is insignificant because the temperature change is small. The initial specific enthalpy of air is $h_0 = 419.40\ \mathrm{kJ/kg}$. Thus, the enthalpy is

$$\begin{align} H_0 &= m \cdot h_0 \\ &= 102.391\ \mathrm{kg} \times 419.40\ \mathrm{kJ/kg} \\ &= 42\,943\ \mathrm{kJ} \end{align}$$

At constant pressure, the change in enthalpy corresponds to the heat transferred to the air:

$$\begin{align} H_1 &= H_0 + \Delta H \\ &= H_0 + Q \\ &= 42\,943\ \mathrm{kJ} + 216\ \mathrm{kJ} \\ &= 43\,159\ \mathrm{kJ} \end{align}$$

Thus, the final specific enthalpy is

$$\begin{align} h_1 &= \frac{H_1}{m} \\ &= \frac{43\,159\ \mathrm{kJ}}{102.391\ \mathrm{kg}} \\ &= 421.51\ \mathrm{kJ/kg} \end{align}$$

At a pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, this specific enthalpy corresponds to a temperature of $T = 22.1\ \mathrm{^\circ C}$.

Note that, at the final temperature of $T = 22.1\ \mathrm{^\circ C}$ and the pressure of $p = 1\ \mathrm{atm} = 101\,325\ \mathrm{Pa}$, the new density of air amounts to $\rho_1 = 1.1960\ \mathrm{kg/m^3}$. Thus, the new volume is

$$\begin{align} V_1 &= \frac{m}{\rho_1} \\ &= \frac{102.391\ \mathrm{kg}}{1.1960\ \mathrm{kg/m^3}} \\ &= 85.6\ \mathrm{m^3} \end{align}$$

Therefore, a small amount of air must escape from the room.

Finally, if the heat from a $60\ \mathrm W$ light bulb really is a problem, you might want to leave the room since the human body generates about $100\ \mathrm W$.

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There is no real answer to this because light bulbs do not have a exact temperature do to the various shapes sizes. Also some are Built with different chemicals or metals. But let's say it is a basic light bulb if you touch it warms Ur hand to burning. There is also light radiation as a factor warming the surfaces it connects with so yes a light bulb warms the standard room. To prove this just take a wind mill and hold it over a lamp the windmill will spin slowly form the rise of hot air.

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A 60 watt equivalent LED bulb uses about 9 watts of power, so we can conclude that an incandescent bulb produces at least 51 watts of heat energy for every 9 watts of light energy.

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  • $\begingroup$ Actually, a 60-watt lamp produces about 60 watts of heat energy, not counting any light that escapes through windows or converted to glucose by house plants. All the light produced, as well as IR, will bounce around the room until it is absorbed and turned to heat. LED and CFL lamps produce just about as much heat as the electricity consumed. $\endgroup$ – DrMoishe Pippik Aug 30 '15 at 5:00
  • $\begingroup$ You are right, DrMoishe. $\endgroup$ – iad22agp Aug 30 '15 at 7:40

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