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 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 calculating an answer close to 30 and im not sure what im doing wrong*

Here's my attempt.

V = 324.8 m^3
Total Energy of Students: 395280 J (in 54min)
mass = 406kg

q=mc(deltaT)

q= 395280J
c= 840 J/kg C
m= 406 kg

delta T = 1.16 C

Tf = 20.0 C

To find the temperature of the room at the end of 54.0 minutes, we can use the formula for heat transfer:

Q = mcΔT

Where:
Q is the heat transferred to the air (in joules),
m is the mass of the air (in kilograms),
c is the specific heat capacity of air (in J/kg*oC), and
ΔT is the change in temperature (in oC).

First, we need to find the mass of the air in the room. Using the given density and the dimensions of the room, we can calculate the volume of the room in cubic meters:

V = length x width x height
= 6.40 m x 14.5 m x 3.50 m
= 325.12 m^3

Since the density is given in grams per cubic centimeter, we need to convert the density to kilograms per cubic meter:

density = 1.25E-3 g/cm^3 = 1.25 kg/m^3

The mass of the air is then calculated by multiplying the volume and density:

m = density x V
= 1.25 kg/m^3 x 325.12 m^3
= 406.4 kg

Next, we calculate the heat transferred to the air using the formula:

Q = mcΔT

Since the initial temperature of the room is 18.8oC and the final temperature is unknown, we can rewrite the formula as:

Q = mc(Tf - Ti)

Substituting the values:

Q = 406.4 kg x 840 J/kg*oC x (Tf - 18.8oC)

The heat output from the 11 students is given in watts, which is equivalent to joules per second (J/s). Therefore, the total heat output over 54.0 minutes (3240 seconds) is calculated as:

Q = Output x time
= (11 students x 122 W) x 3240 s

Now we can equate the two expressions for Q:

(11 students x 122 W) x 3240 s = 406.4 kg x 840 J/kg*oC x (Tf - 18.8oC)

After simplifying and rearranging the equation:

(Tf - 18.8oC) = [(11 students x 122 W x 3240 s) / (406.4 kg x 840 J/kg*oC)]

Solving for Tf:

Tf = ([(11 students x 122 W x 3240 s) / (406.4 kg x 840 J/kg*oC)]) + 18.8oC

Plugging in the given values and performing the calculation will give you the final temperature of the room in degrees Celsius.