Q is the amount of energy added (joules) M is the mass of the fluid (kg) C (subs
ID: 2128151 • Letter: Q
Question
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
Specified Information:
Number of bulbs:6
bulb power: 60 W
bulb percent power loss as heat:90%
bulbs on time:3 hrs
garage air volume: 5760 ftcubed
air density: 1.2 kg/mcubed
Air heat capacity: 1000 joules/Kg K
_______________________________________
Calculated Information:
Total Bulb Power? ......W
Total bulb power lost as heat?.......W
Total bulb power lost as heat?......joules/second
Total bulb energy lost as heat?..... joules
Garage Air Volume?.............m cubed
Garage Air mass?........Kg
Temperature Change?.....K
Explanation / Answer
E = 0.90x6x60 Joules per second(watt) assuming 90% electrical energy consumed is transmitted as heat.
How would I figure out the Total Bulb Energy Lost, ET as Heat in terms of joules?
ET = 3x3600x0.90x6x60 Jules
How would I calculate the Temperature Change, dT in Kelvins?
dT = 3x3600x0.90x6x60/{1.2x[(24x12x2.54/100)^
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