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.(adsbygoogle = window.adsbygoogle || []).push({});

2. Relevant equations

m = 2CvQ/(3RΔTcv)

ρ=m/v

dQ = mcdT

3. 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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Compute Heat capacity

**Physics Forums | Science Articles, Homework Help, Discussion**