The latent heat of vaporization of water is 2240 J. If the work done in the process of vaporization of 1 gram is 168 J, then increase in internal energy is

$\begin{array}{rcl} & & \\ & & first law of thermodynamics is \text{dQ=dU+dW}\\ & & dQ is heat energy; dU is internal energy; dW is work done\\ given mass & =& 1gram; heat given to the system=dQ=2240J; work done by the system=168J\\ & \Rightarrow & \text{dU=dQ-dW}\\ & & substitute the given values in equation dU\\ & \Rightarrow & \mathrm{dU}=2240-168\\ increase in internal energy of system& =& 2072 \mathrm{J}.\end{array}$