Before we solve for the final temperature, the initial temperature should be given. Suppose this is at room temperature, 25°C. The solution is as follows.
8.8 g(1/Molar mass of benzene)(Heat of combustion of benzene) = 5,691 g water(Heat capacity of water)(Tfinal - Tinitial)
Heat from benzene = - Heat from water
8.8 g(1/78 g/mol)(-3,271 kJ/mol) = -(5,691 g)(4.816 J/g·°C)(1 kJ/1000 J)(Tfinal - 25°C)
<em>Tfinal = 38.46°C</em>
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.
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