Recent content by Thomas_Kellner

It’s not warming significantly. But how is that going to help calculating the final temperature of the air inside the glass?

The mass is about 400 g. So the increase of temperature would be : 0,3 KJ/0,4 kg*0,72 KJ/kg*K = 1,04 K

1) The flame extinguishes itself after 10 seconds. 2) It’s relatively thick which is why we assume the system is closed. You mentioned radiation, does that mean a more appropriate way of calculating the final temperature is via the Stefan Boltzmann law? Also, what role does convection play? 3)...

So basically, only about 0,00027 KJ of warmth is transferred to the air in the system and the rest is stored in the paraffin wax and glass? How is that possible if the specific heat capacity for paraffin wax is about 2,1 KJ/kg*K and for glass is about 0,7 KJ/kg*K? Also, the warmth given off by...

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...