m₁ = 26.6 g, the mass of aluminum
T₁ = 100.4 °C, the temperature of the aluminum
c₁ = 0.903 J/(g-°C), the specific heat of aluminum
m₂ = 100.6 g, the mass of water
T₂ = 21.5 °C, the temperature of water
The specific heat of water is c₂ = 4.184 J/(g-C)
Let T = the final temperature of the water and aluminum.
Calculate heat loss of the aluminum.
Q₁ = (26.6 g)*(0.903 J/(g-C))*(100.4-T C) = 24.1098(100.4 - T) J
Calculate heat gained by the water
Q₂ = (100.6 g)*(4.184 J/(g-C))*(T - 21.5 C) = 420.1904(T - 21.5) J
Conservation of energy requires that Q₁ = Q₂.
420.1904(T - 21.5) = 24.1098(100.4 - T)
444.3T = 1.1455 x 10⁴
T = 25.782 °C
the dissolution of polymer can be thermodynamically described in 2 steps.
hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.
on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.
thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.