Where: Q is the amount of energy added (joules) M is the mass of the fluid (kg)

Question

Where:
Q is the amount of energy added (joules)
M is the mass of the fluid (kg)
C (subscript p) is the heat capacity of the fluid (joules/kg/K), and
Delta T is the change in temperature (K or C degrees)

A garage (24 ft x 24 ft x 10 ft) is illuminated by (6) 60 watt incandescent bulbs. It is estimated that 90% of the energy to an incandescent bulb is dissipated as heat. If the bulbs are left on for 3 hours, how much would the temperature in the garage increase because of the light bulbs (assuming no energy losses).

Here is some potentially useful information:
Air density (approximate): 1.2 kg/m cubed
Air heat capacity (approximate): 1000 joules/kg?K
3.28 ft = 1 m

Explanation / Answer

Electrical energy, E = Power * time = 6 * 60 * 3 * 3600 J Heat energy, Q = 90% of 6 * 60 * 3 * 3600 = 3.88 x 10^6 J Mass, M = Density * Volume = 1.2 * 24 * 24 * 10 * ( 1 / 3.28 ^3 ) = 195.9 kg Heat capacity = 1000 J/kg/K Rise in temperature = ? Heat , Q = Mass * Heat capacity * Rise in temp 3.88 x 10^6 = 195.9 * 1000 * Rise in temp Rise in temp = 19.85 K (or) 19.85 deg celsius

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