Answer
$Q=20.06MJ/kg$
Work Step by Step
The energy balance for the water system is:
$Q=\Delta U=mc_v\Delta T$
Given $m=3\ kg, c_v=4.18\ kJ/kg.K,\ \Delta T=3.2K$:
$Q=40.13\ kJ$
Hence the energy content for the 2g sample is:
$q=Q/m=20.06MJ/kg$
The error estimate is given by the change in internal energy of the air:
$\Delta U_{air}=mc_v\Delta T$
With $m=0.102kg,\ c_v=0.718\ kJ/kg.K,\ \Delta T=3.2K$:
$\Delta U_{air}=0.23\ kJ$
