How many grams of steam at 100 Celsius would be required to raise the temp. of 35.8 g solid benzene, from 5.5 Celsius to 45.0 Celsius???
[mass benzene x heat fusion benzene] + [mass benzene x specific heat benzene x (Tfinal-Tinitial)] + [mass steam x heat vaporization of water] + [mass water@100 x specific heat water x (Tfinal-Tinitial)] = 0
Check my thinking.
To find out how many grams of steam are required to raise the temperature of solid benzene, you need to use the specific heat capacity of benzene and the heat of vaporization of water.
1. Calculate the heat absorbed by the benzene to raise its temperature from 5.5°C to 45.0°C using the specific heat formula:
q = m * c * ΔT
Where:
q = heat absorbed (in Joules)
m = mass of substance (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
The specific heat capacity of benzene is 1.74 J/g°C. So:
q = 35.8 g * 1.74 J/g°C * (45.0°C - 5.5°C)
2. Calculate the heat required to vaporize the corresponding amount of water (steam) at 100°C from the heat absorbed in step 1 using the heat of vaporization of water.
The heat of vaporization of water is approximately 40.7 kJ/mol or 40.7 J/g.
Convert the heat absorbed in step 1 from Joules to grams:
Heat required (in g) = q / 40.7 J/g
This will give you the amount of water (steam) in grams required to raise the temperature of benzene.
Note: It is important to mention that this calculation assumes complete heat transfer between the steam and benzene without any losses. In reality, there may be some energy losses due to heat transfer mechanisms.