I have more information to a problem I asked about earlier. I continue to get an

Question

I have more information to a problem I asked about earlier. I continue to get an answer of 18 degrees Kelvin which is different from an answer I received earlier. It is my fault since I did not give the entire question previously.

Here is the question:
A garage (24ftx24ftx10ft) is illuminated by (6) 60W incandescent bulbs. It is estimated that 90% of the energy to an incandescent bulb is dissipated as heat. If the bulbs are on for 3 hours, how much would the temperature in the garage increase because of the lightbulbs (assuming no energy losses).
Information Given:

Formula Q=MCpDeltaT
Q is the amount of energy added (joules)
M is the mass of the fluid (kg)
Cp is the heat capacity of the fluid (joules/kg/K)
Delta T is the change in temperature (K or °C)

Number of Lightbulbs: 6
Power From Each Lightbulb (Joules/second): 60
Bulb % Power Loss as Heat: 90%
Amount of Time Bulbs in "On" Position (Hrs): 3
Time Bulbs On (Seconds): 10800
Garage Air Volume (ft3): 5760
Air Density (kg/m³): 1.2
Air Heat Capacity (Joules/kg/K): 1000


Three Hrs of Power from 1 Bulb (watts): ?
Three Hrs of Power All 6 Bulbs (watts): ?
Total Bulb Power Lost as Heat (joules/second): ?
Total Bulb Energy Lost as Heat (joules): ?
Garage Air Volume (m³): ?
Garage Air Mass (kg): ?
Temperature Change (K): ?

Explanation / Answer

Formula Q=MCpDeltaT Q is the amount of energy added (joules) M is the mass of the fluid (kg) Cp is the heat capacity of the fluid (joules/kg/K) Delta T is the change in temperature (K or °C) Number of Lightbulbs: 6 Power From Each Lightbulb (Joules/second): 60 Bulb % Power Loss as Heat: 90% Amount of Time Bulbs in "On" Position (Hrs): 3 Time Bulbs On (Seconds): 10800 Garage Air Volume (ft3): 5760 Air Density (kg/m³): 1.2 Air Heat Capacity (Joules/kg/K): 1000 Three Hrs of Energy from 1 Bulb = Power From Each Lightbulb * time = 60 J/s * 10800 s = 648000 J Three Hrs of Energy All 6 Bulbs = 648000*6 = 3888000 J Total Bulb Energy Lost as Heat : 90 * 3888000 / 100 = 3499200 J Garage Air Volume = 56.56 m^3 (1 ft = 0.3048 m) Garage Air Mass = density * volume = 1.2 * 56.56 = 67.872 kg Temperature Change = Q/MCp = 3499200 J /67.872 kg*1000 Joules/kg/K = 51.55 K

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