Answer:
There was 450.068g of water in the pot.
Explanation:
Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L
Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s
Let m = x g be the weight of water in the pot.
Energy required to vaporise water = mL = 2260x
Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x
Total energy required =
Hence, there was 450.068g of water in the pot.
Answer:
At -13 , the gas would occupy 1.30L at 210.0 kPa.
Explanation:
Let's assume the gas behaves ideally.
As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-
where and
are initial and final pressure respectively.
and
are initial and final volume respectively.
and
are initial and final temperature in kelvin scale respectively.
Here ,
,
,
and
Hence
So at -13 , the gas would occupy 1.30L at 210.0 kPa.