- #1

Thomas_Kellner

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- Homework Statement
- Assume you have a candle inside a drinking glass (cylinder turned upside down) that burns for

10 s.

You know the following values.

Air density = 1.2 kg/m^3

Diameter of the cylinder = 6,5 cm

Height of the cylinder = 6 cm

Initial temperature = 293,15 K

Heating power of the candle = 30 W

Time = 10 s

Specific heat capacity of air = 1 KJ/kg*K

Calculate the final temperature of the air inside the glass.

- Relevant Equations
- c = Q/m*ΔT

Q = P*t

Looking at the given values, I thought the specific heat formula could be used to calculate the final temperature of the air: c = Q/mΔT. Since the final temperature is the sum of the initial temperature and the change in temperature, the formula can be rearranged to ΔT = Q/m*c.

Q = 30 W * 10 s = 300 J

V = A * h = (

ΔT = Q/m*c = 0,3 KJ/(0,00024 kg * 1KJ/kg*K) = 1250 K

Final temperature = 1250 K + 293,15 K = 1543,15 K

This number seems physically impossible, so I conducted the experiment with an initial temperature of 293,15 K, the same air volume and a normal tealight that burned for approximately 10 s. The oven thermometer inside showed a final temperature of about 304,15 K, meaning ΔT = 11K. What did I do wrong?

Q = 30 W * 10 s = 300 J

V = A * h = (

*π * r^2) * 6 cm = (**π * 3,25 cm^2) * 6 cm = 33,2 cm^2 * 6 cm = 199 cm^3*

m = V * p = 0,000199 m^3 * 1,2 kg/m^3 = 0,00024 kg

c = 1 KJ/kg*Km = V * p = 0,000199 m^3 * 1,2 kg/m^3 = 0,00024 kg

c = 1 KJ/kg*K

ΔT = Q/m*c = 0,3 KJ/(0,00024 kg * 1KJ/kg*K) = 1250 K

Final temperature = 1250 K + 293,15 K = 1543,15 K

This number seems physically impossible, so I conducted the experiment with an initial temperature of 293,15 K, the same air volume and a normal tealight that burned for approximately 10 s. The oven thermometer inside showed a final temperature of about 304,15 K, meaning ΔT = 11K. What did I do wrong?