How Much Air Can Be Heated with the Same Energy Used to Warm Water in a Kettle?

[Firben]

You warm 1.00 kg of water at a constant volume of 1.00 L from 20.0 degrees C to 30.0 degrees C in a kettle. For the same amount of heat, how many kilo grams of 20.0 degree C air would you be able to warm to 30.0 degree C? What volume would this air occupy at 20.0 degrees C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N2.

Homework Equations

m = 2CvQ/(3RΔTcv)

ρ=m/v

dQ = mcdT

The Attempt at a Solution

Q = (1.00kg)(4190 J/mol*K)(303K-293K) = 41900 J

c(air) = 1.01 * 10^3 J/kg*K
V= 1.00 L
ρ(air) =1.2929 kg/m^-3
m(air) = ρ(air)*V <=>
m(air) = (1.2929 kg/m^-3)(1.00 L) = 1.2929 kg
C(air) = (1.01*10^3 J/kg*K)(1.2929 kg) = 1305.829 J/mol*K

R = 8.314 J/mol*K
cv = 4190 J/mol*K

the answer should be 5.65 kg

 

[Borek]

Hint: you ignored V=const part.

 

[Firben]

Is this formula right m = 2CvQ/(3RΔTcv) ?

 

[Borek]

Argh, what was I thinking. You used wrong heat capacity value, but not because of the constant volume, you just have your units wrong. Sigh.

Is this formula right m = 2CvQ/(3RΔTcv) ?

Check your units. Is the result in kg?

 

[asteriatic]

of air at 20 degrees C

To compute the heat capacity, we can use the equation Q = mcdT, where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and dT is the change in temperature.

In this case, we have a mass of 1.00 kg of water and a change in temperature of 10 degrees C, giving us a value of 41900 J for Q. We can then rearrange the equation to solve for c, the specific heat capacity of water, which is equal to 41900 J / (1.00 kg * 10 K) = 4190 J/kg*K.

To find the amount of air that can be warmed to the same temperature with the same amount of heat, we can use the equation m = (2CvQ)/(3RΔTcv), where m is the mass of the substance, Cv is the specific heat capacity at constant volume, and R is the gas constant. We can also use the ideal gas law, PV = nRT, to find the volume of air at 20 degrees C and 1.00 atm.

Using the given specific heat capacity of air (1.01*10^3 J/kg*K) and the known values for R and Cv, we can solve for m, which gives us a value of 5.65 kg of air. This means that for the same amount of heat, we can warm 5.65 kg of air from 20 degrees C to 30 degrees C.

To find the volume of this air at 20 degrees C and 1.00 atm, we can use the ideal gas law. Rearranging the equation to solve for V, the volume, we get V = nRT/P. Using the known values for n (number of moles), R, and P, we can calculate the volume to be 480.91 L.

However, in this calculation, we made the simplifying assumption that air is 100% N2. In reality, air is composed of various gases, so this calculation will only give us an approximate answer.