1. 408.4 J
The work done by a gas is given by:
[tex]W=p\Delta V[/tex]
where
p is the gas pressure
[tex]\Delta V[/tex] is the change in volume of the gas
In this problem,
[tex]p=1.01\cdot 10^5 Pa[/tex] (atmospheric pressure)
[tex]\Delta V=4048.3 cm^3 - 4.5 cm^3 =4043.8 cm^3 = 4043.8 \cdot 10^{-6}m^3[/tex] is the change in volume
So, the work done is
[tex]W=(1.01\cdot 10^5 Pa)(4043.8 \cdot 10^{-6}m^3)=408.4 J[/tex]
2. 10170 J
The amount of heat added to the water to completely boil it is equal to the latent heat of vaporization:
[tex]Q = m \lambda_v[/tex]
where
m is the mass of the water
[tex]\lambda_v = 2.26\cdot 10^6 J/kg[/tex] is the specific latent heat of vaporization
The initial volume of water is
[tex]V_i = 4.5 cm^3 = 4.5\cdot 10^{-6}m^3[/tex]
and the water density is
[tex]\rho = 1000 kg/m^3[/tex]
So the water mass is
[tex]m=\rho V_i = (1000 kg/m^3)(4.5\cdot 10^{-6}m^3)=4.5\cdot 10^{-3}kg[/tex]
So, the amount of heat added to the water is
[tex]Q=(4.5\cdot 10^{-3}kg)(2.26\cdot 10^6 J/kg)=10,170 J[/tex]