An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 6635 J. What is the specific heat of the gas?
Well, first off, I must say, that gas really heated things up! Anyway, let's get down to the specifics.
To find the specific heat of the gas, we can use the equation:
q = mCΔT
where q is the heat transferred, m is the mass of the gas, C is the specific heat, and ΔT is the change in temperature.
Now, we know that the work done by the system is 346 J, which means the heat transferred is equal to the change in internal energy plus the work done:
q = ΔE + W
Plugging in the values we have, we get:
q = 6635 J + 346 J
q = 6981 J
Since the mass is given as 80.0 grams, which is the same as 0.080 kg, we can rearrange the equation to solve for C:
C = q / (mΔT)
C = 6981 J / (0.080 kg × (225 °C - 25 °C))
C = 6981 J / (0.080 kg × 200 °C)
C ≈ 436.31 J/(kg °C)
So, the specific heat of the gas is approximately 436.31 J/(kg °C).
To find the specific heat of the gas, we need to use the formula:
Q = mcΔT
Where:
Q = amount of heat absorbed or released
m = mass of the gas
c = specific heat capacity of the gas
ΔT = change in temperature
In this case, the heat absorbed is equal to the sum of the work done by the system and the change in internal energy:
Q = 346 J + 6635 J
Q = 6981 J
Now, we can rearrange the formula to solve for the specific heat capacity (c):
c = Q/(mΔT)
Plugging in the values:
c = 6981 J / (80.0 g * (225 °C - 25 °C))
Simplifying the equation:
c = 6981 J / (80.0 g * 200 °C)
Finally, we can calculate the specific heat capacity:
c = 0.436 J/(g·°C)
Therefore, the specific heat of the gas is approximately 0.436 J/(g·°C).
To find the specific heat of the gas, we can use the formula:
ΔQ = m * c * ΔT
where:
ΔQ is the heat absorbed or released by the gas
m is the mass of the gas
c is the specific heat of the gas
ΔT is the change in temperature
In this case, the heat absorbed by the gas (ΔQ) is the sum of the work done by the system (346 J) and the increase in internal energy (6635 J):
ΔQ = 346 J + 6635 J = 6981 J
The mass of the gas (m) is given as 80.0 grams, and the change in temperature (ΔT) is the final temperature (225 °C) minus the initial temperature (25 °C):
ΔT = 225 °C - 25 °C = 200 °C
Now we can rearrange the formula to solve for the specific heat (c):
c = ΔQ / (m * ΔT)
Plugging in the known values:
c = 6981 J / (80.0 g * 200 °C)
Before we can calculate the specific heat, we need to convert the mass from grams to kilograms and the temperature from Celsius to Kelvin:
m = 80.0 g = 0.080 kg
ΔT = 200 °C = 200 K
Now we can calculate the specific heat:
c = 6981 J / (0.080 kg * 200 K)
c ≈ 43.7 J/(kg·K)
Therefore, the specific heat of the gas is approximately 43.7 J/(kg·K).