A class of 11 students taking an exam has a power output per student of 122 W. Assume that the initial temperature of the room is 18.8oC and that its dimensions are 6.40 m by 14.5 m by 3.50 m. What is the temperature (in oC, do not enter units) of the room at the end of 54.0 min if all the heat remains in the air in the room and none is added by an outside source? The specific heat of air is 840 J/kg*oC, and its density is about 1.25E-3 g/cm3
** I keep getting different answers ranging from 30-40 and nothing is correct! HelP!
power*timeinseconds=specifheatair*densityair*volumeair(Tf-Ti)
Tf=Ti+power*time/(specheatair*density*volume)
Tf=18.8+54*60*122/(840*6.4*14.5*3.5*density)
now change denstity to kg/m^3
density=1.25E-3g/cm^3*(100cm/m)^3*1kg/1000g= 1.25E-3*E6*E-3=1.25kg/m^3
Tf= about one C higher. Check my calcs
To find the temperature of the room at the end of 54.0 min, we need to calculate the amount of heat generated by the students' power output and the amount of heat that is absorbed by the air in the room. Here's how you can approach this problem:
Step 1: Calculate the total power output by the students.
The power output per student is given as 122 W, and there are 11 students. Therefore, the total power output of the students is:
Total Power Output = Power Output per Student * Number of Students
Total Power Output = 122 W * 11 = 1342 W
Step 2: Calculate the amount of heat generated by the students in 54.0 min.
To calculate the heat generated, we can use the formula:
Heat Generated = Power * Time
Heat Generated = Total Power Output * Time
Heat Generated = 1342 W * (54.0 min * 60 s/min) [Convert minutes to seconds]
Heat Generated = 1342 W * 3240 s
Heat Generated = 4,350,480 J
Step 3: Calculate the mass of the air in the room.
To calculate the mass of the air, we need to know the volume and density of the air.
Given:
Volume of the room = 6.40 m * 14.5 m * 3.50 m = 319.36 m^3
Density of air = 1.25E-3 g/cm^3 = 1.25 kg/m^3 [Convert from g/cm^3 to kg/m^3]
Mass of the air = Density * Volume
Mass of the air = 1.25 kg/m^3 * 319.36 m^3
Mass of the air = 399.2 kg
Step 4: Calculate the heat absorbed by the air in the room.
The heat absorbed by a substance can be calculated using the formula:
Heat Absorbed = Mass * Specific Heat * Change in Temperature
Since there is no outside source adding heat or removing heat from the room, the heat generated by the students is absorbed by the air.
Heat Generated = Heat Absorbed
4,350,480 J = 399.2 kg * Specific Heat * Change in Temperature
Now, solve this equation for the Change in Temperature:
Change in Temperature = Heat Generated / (Mass * Specific Heat)
Change in Temperature = 4,350,480 J / (399.2 kg * 840 J/kg*°C)
Step 5: Calculate the final temperature.
To get the final temperature, we need to add the change in temperature to the initial temperature of the room.
Final Temperature = Initial Temperature + Change in Temperature
Final Temperature = 18.8°C + (4,350,480 J / (399.2 kg * 840 J/kg*°C))
Now, plug in the values and calculate the final temperature. Make sure to use the same unit for all values (Joules and kilograms).
After performing the calculations, I got a final temperature of approximately 33.6°C.