A class of 11 students taking an exam has a power output per student of 124 W. Assume that the initial temperature of the room is 18.7°C and that its dimensions are 6.00 m by 14.5 m by 3.40 m. What is the temperature (in °C, do not enter units) of the room at the end of 55.0 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 832 J/kg*°C, and its density is about 1.34E-3g/ cm3.
Well, it looks like the students in that class are really "heating" things up with their exam! Let's see how hot it gets in there.
To find the change in temperature of the room, we need to calculate the amount of heat transferred. The formula for heat transfer is Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the mass of the air in the room. The volume of the room is given as 6.00 m * 14.5 m * 3.40 m = 293.4 m^3. To convert this to grams, we multiply by the density of air: 293.4 m^3 * 1.34E-3 g/cm^3 = 0.393 g.
Now, let's calculate the heat transferred. The power output per student is given as 124 W. To find the total power output, we multiply this by the number of students: 124 W * 11 = 1364 W.
Since power is defined as the rate of energy transfer, we can calculate the energy transferred over 55.0 min using the formula E = Pt, where E is the energy, P is the power, and t is the time: 1364 W * 55.0 min * 60 s/min = 4496400 J.
Now that we have the energy transferred and the mass of air, we can use the formula Q = mcΔT to find the change in temperature. Rearranging the formula gives us ΔT = Q / (mc). Plugging in the values, we get ΔT = 4496400 J / (0.393 g * 832 J/kg*°C) ≈ 677.04 °C.
So, at the end of 55.0 min, the temperature of the room would be approximately 677.04 °C. But hold on a second, that's hotter than a jalapeno pepper! Something doesn't quite add up here. It seems like we may have made some calculations errors along the way. Let's go back and double-check our work to get the correct answer.
To find the temperature of the room at the end of 55.0 min, we need to calculate the change in temperature using the formula:
ΔQ = mcΔT
Where:
ΔQ is the change in heat energy
m is the mass of the air in the room
c is the specific heat of air
ΔT is the change in temperature
Step 1: Calculate the mass of the air in the room
To find the mass of the air in the room, we need to know the volume of the room and the density of air.
Given:
Volume of the room = 6.00 m x 14.5 m x 3.40 m = 295.20 m^3
Density of air = 1.34E-3 g/cm^3
Converting density from g/cm^3 to kg/m^3:
1.34E-3 g/cm^3 * (1 kg / 1000 g) * (1000000 cm^3 / 1 m^3) = 1.34 kg/m^3
Mass of the air in the room = volume x density
= 295.20 m^3 x 1.34 kg/m^3
= 395.23 kg
Step 2: Calculate the change in heat energy (ΔQ)
To find the change in heat energy, we need to know the power output per student and the time.
Given:
Power output per student = 124 W
Number of students = 11
Total power output in the room = 124 W x 11
= 1364 W
Total energy supplied in 55.0 min = power output x time
= 1364 W x (55.0 min x 60 s/min)
= 4478400 J
Step 3: Calculate the change in temperature (ΔT)
We can now use the equation ΔQ = mcΔT to find the change in temperature.
ΔQ = mcΔT
4478400 J = 395.23 kg x 832 J/kg*°C x ΔT
Solving for ΔT:
ΔT = (4478400 J) / (395.23 kg x 832 J/kg*°C)
ΔT ≈ 6.12°C
Step 4: Find the final temperature
To find the final temperature, we need to add the change in temperature to the initial temperature.
Final temperature = Initial temperature + ΔT
Final temperature = 18.7°C + 6.12°C
Final temperature ≈ 24.82°C
Therefore, the temperature of the room at the end of 55.0 min is approximately 24.82°C.
To find the change in temperature of the room after 55.0 minutes, we can use the formula:
Q = mcΔT
Where:
Q = heat energy transferred (in joules)
m = mass of the air in the room (in kilograms)
c = specific heat of air (in joules per kilogram-degree Celsius)
ΔT = change in temperature (in degrees Celsius)
First, let's calculate the mass of the air in the room:
Density (ρ) = mass/volume
Given that the density of air is 1.34E-3 g/cm³, we can convert it to kg/m³ by dividing it by 1000:
Density (ρ) = 1.34E-3 g/cm³ / 1000 g/kg = 1.34E-6 kg/cm³
Now, we can calculate the mass by multiplying the density by the volume of the room:
Volume (V) = length × width × height
V = 6.00 m × 14.5 m × 3.40 m = 294.6 m³
Mass (m) = density × volume
m = 1.34E-6 kg/cm³ × 294.6 m³ = 0.000394 kg
Now, we can calculate the heat energy (Q) transferred:
Q = power × time
Q = (power per student × number of students) × time
Q = (124 W × 11) × (55.0 min × 60 s/min)
Q = 918,120 J
Next, we can rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
Substituting the values:
ΔT = 918,120 J / (0.000394 kg × 832 J/kg*°C)
Now, we can solve for ΔT:
ΔT ≈ 135.54 °C
Therefore, the temperature of the room at the end of 55.0 minutes would be approximately 135.54 °C.