60) A class of 10 students taking an exam has a power output per student of about 200W. Assume the initial temperature of the room is 20 C and that its dimensions are 6.0 m by 15.0 m by 3.0 m. What is the temperature of the room at the end of 1.0 h if all the energy remains in the air in the room and none is added by an outside source? The specific heat is 837 J/kg C, and its density is about 1.3 x 10^-3 g/cm^3

I just need the work shown
The answer is 44.5 C (= 112 F)

The heat generated by the students at the end of one hour is

10*(200 J/s)*3600 s = ___ Joules

Divide that by the heat capacity of the air in the room,
(Volume)*(air density)*Specific heat)

The ratio will be the temperature rise.

The high value is the result of erroneous assumptions. In a real case, most of the added heat would be lost to the walls and objects in the room.

You do the calculations

I did as you said. The only thing I did not see you add to the equation though is the initial temperature of the room. I don't know if I did it right because after your instructions I just added that 20 C to the final answer and it gave me 44.5! I don't know if it's right but it gave me the correct answer.

To find the final temperature of the room at the end of 1.0 hour, we need to calculate the change in thermal energy of the air in the room.

Step 1: Calculate the initial thermal energy of the air in the room.
The formula for thermal energy is:
Thermal energy = mass * specific heat * change in temperature

The mass of the air in the room can be calculated using the density formula:
mass = density * volume

Given that the density is 1.3 x 10^-3 g/cm^3 and the volume of the room is 6.0 m * 15.0 m * 3.0 m, we need to convert the volume to cm^3:
volume = 6.0 m * 100 cm/m * 15.0 m * 100 cm/m * 3.0 m * 100 cm/m = 2.7 x 10^6 cm^3

mass = density * volume = 1.3 x 10^-3 g/cm^3 * 2.7 x 10^6 cm^3 = 3.51 kg

The change in temperature is given as the difference between the final and initial temperatures:
change in temperature = final temperature - initial temperature = T - 20 C

The initial thermal energy can be calculated as:
initial thermal energy = mass * specific heat * change in temperature = 3.51 kg * 837 J/kg C * (T - 20 C)

Step 2: Calculate the total power output of the class.
The power output per student is given as 200W, and we have a class of 10 students.
The total power output of the class is given by:
total power output = power output per student * number of students = 200W * 10 = 2000W

Step 3: Calculate the total energy output of the class over 1 hour.
The total energy output can be calculated using the power formula:
total energy output = power * time = 2000W * 1.0 hour = 2000 J

Step 4: Equate the total energy output to the initial thermal energy to solve for the final temperature.
2000 J = 3.51 kg * 837 J/kg C * (T - 20 C)

Now, let's solve for T:
2000 J / (3.51 kg * 837 J/kg C) = T - 20 C
T - 20 C = 0.709
T = 0.709 + 20 C
T ≈ 20.709 C

Therefore, the temperature of the room at the end of 1.0 hour is approximately 20.709 degrees Celsius.

To find the temperature of the room at the end of 1.0 hour, we need to calculate the energy transferred to the air in the room.

First, let's find the total power output of the class. Since there are 10 students and each one has a power output of 200W, the total power output is:

Total Power Output = Number of Students × Power Output per student
Total Power Output = 10 × 200 W
Total Power Output = 2000 W

Now, we need to convert this power to energy. Energy is given by the equation:

Energy = Power × Time

In this case, the time is 1.0 hour. However, we need to convert this to seconds since all the other units are in the SI system. There are 3600 seconds in 1 hour.

Time = 1.0 hour × 3600 seconds/hour
Time = 3600 seconds

Now we can find the energy transferred to the air in the room:

Energy = Total Power Output × Time
Energy = 2000 W × 3600 seconds
Energy = 7,200,000 J

Next, we need to calculate the mass of the air in the room. To do this, we'll use the formula:

Density = Mass / Volume

We have the density of the air, which is given as 1.3 x 10^-3 g/cm^3. Let's convert this to kg/m^3. Since there are 100 cm in a meter and 1000 g in a kg, we can multiply the density by 1000 to convert g/cm^3 to kg/m^3.

Density = 1.3 x 10^-3 g/cm^3 × 1000 kg/g
Density = 1.3 kg/m^3

Next, we need to find the volume of the room. The volume is given by the formula:

Volume = Length × Width × Height
Volume = 6.0 m × 15.0 m × 3.0 m
Volume = 270 m^3

Now we can find the mass of the air:

Mass = Density × Volume
Mass = 1.3 kg/m^3 × 270 m^3
Mass = 351 kg

Finally, we can calculate the change in temperature using the equation:

Energy = Mass × Specific Heat × Change in Temperature

Rearranging the equation to solve for the change in temperature, we get:

Change in Temperature = Energy / (Mass × Specific Heat)

Plugging in the values we have:

Change in Temperature = 7,200,000 J / (351 kg × 837 J/kg °C)
Change in Temperature ≈ 26.68 °C

Since the initial temperature of the room was 20 °C, we can find the final temperature by adding the change in temperature to the initial temperature:

Final Temperature = Initial Temperature + Change in Temperature
Final Temperature = 20 °C + 26.68 °C
Final Temperature = 46.68 °C

Rounding to one decimal place, the temperature at the end of 1.0 hour would be approximately 46.7 °C, which is equivalent to 112 °F.