Quizzes & Puzzles9 mins ago
Thermodynamics
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Consider a classroom of dimension5m*10m*3m at temperature 20 celsuis and 1 atm pressure .There are 50 people in the room each loosing energy at the average of 150 watt. Assuming that the walls ,ceiling,floors and furniture are perfectly insulated and none of them absorbs heat. How much time will be needed in rising the temperature of air to body temperature (37 celsuis).
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For more on marking an answer as the "Best Answer", please visit our FAQ.That's amazing, I was wondering the same thing too.
Sounds like a homework question.
Thinking back 35 years to physics lessons i recall PV/T is constant
if V is fixed then P/t= constant
Original P/t is 1/ 293 (where temp is in degrees kelvin)
New t= 273+37= 310
so new P=310/293= 1.058
ow need to work out the time taken. I think it's something to do with Watts-joules/second
Total watts = 50 x 150= 7500
Sure you can finish it off.
In this case change in t is 15 degrees
Sounds like a homework question.
Thinking back 35 years to physics lessons i recall PV/T is constant
if V is fixed then P/t= constant
Original P/t is 1/ 293 (where temp is in degrees kelvin)
New t= 273+37= 310
so new P=310/293= 1.058
ow need to work out the time taken. I think it's something to do with Watts-joules/second
Total watts = 50 x 150= 7500
Sure you can finish it off.
In this case change in t is 15 degrees
heat required (kJ) = mass (kg) x Temp difference x heat capacity of air
Density of air is 1.2 kg/m3 so use this to work out the mass of the air in the room.
The temp difference is 17 K.
The heat capacity of air is 1.012 kJ/kg. K
Do the sums and that will give you how many kJ you need
The heat input is 7 500 W which is 7.5 kW or 7.5 kJ/s
Dividing the kJ by the kJ/s will give you the answer in s
Density of air is 1.2 kg/m3 so use this to work out the mass of the air in the room.
The temp difference is 17 K.
The heat capacity of air is 1.012 kJ/kg. K
Do the sums and that will give you how many kJ you need
The heat input is 7 500 W which is 7.5 kW or 7.5 kJ/s
Dividing the kJ by the kJ/s will give you the answer in s