Imagine you\'ve been walking outside on a cold winter\'s day. When you arrive ho

Question

Imagine you've been walking outside on a cold winter's day. When you arrive home at your studio apartment, you realize that you left a window open and your 5.0m

Imagine you've been walking outside on a cold winter's day. When you arrive home at your studio apartment, you realize that you left a window open and your 5.0m times 5.0m times 3.0m room is only slightly warmer than the outside. You turn on your 1.0-kW space heater right away and wait impatiently for the room to warm up. In this problem, make the following assumptions: The entire 1.0kW=1.0 times 10 3 J/s output of the space heater goes into warming the air in the room. The air in the room is an ideal gas with five degrees of freedom per particle (three translational degrees of freedom and two rotational degrees of freedom-about right for nitrogen and oxygen). The air in the room is at a constant pressure of 1.00 atm . At room temperature (22 ? C= 295 K ) and atmospheric pressure, 1.00 mole of air fills a volume of 24.2 L . This is slightly larger than the volume of air at standard temperature and pressure, because room temperature is hotter than 0 ? C How long will it be before the heater warms the air in the room by 10 ? C ?

Explanation / Answer

Volume of the room is V = 5.0m * 5.0m * 3.00m = 75 m^3 Number of moles is n = 75 m^3 * 1000 liters / m^3 * ( 1 mole / 23 liters ) = 3261 moles constant pressure is C_P = 29.06 J / K mole Q = n C_p * T = 3261 moles * 29.06 J / K mole * 10 K = 948 *10^3 J time t = Q / P = 948 *10^3 J / 1.00*10^3 W = 948 s = 15.8 minutes    Volume of the room is V = 5.0m * 5.0m * 3.00m = 75 m^3 Number of moles is n = 75 m^3 * 1000 liters / m^3 * ( 1 mole / 23 liters ) = 3261 moles constant pressure is C_P = 29.06 J / K mole Q = n C_p * T = 3261 moles * 29.06 J / K mole * 10 K = 948 *10^3 J time t = Q / P = 948 *10^3 J / 1.00*10^3 W = 948 s = 15.8 minutes   

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