Answer
$1.34~kg$ of water was in the vessel.
Work Step by Step
The increase in internal energy is equal to the initial gravitational potential energy of the water at a height of 100 meters:
$\Delta E = mgh$
$\Delta E = (1.00~kg)(9.80~m/s^2)(100~m)$
$\Delta E = 980~J$
The internal energy increases by $980~J$
We can find the mass $m$ of water that was in the vessel:
$Q = (m+1.00~kg)~c~\Delta T$
$(m+1.00~kg) = \frac{Q}{c~\Delta T}$
$m = \frac{Q}{c~\Delta T} - 1.00~kg$
$m = \frac{980~J}{(4186~J/kg~C^{\circ})(0.100~C^{\circ})} - 1.00~kg$
$m = 1.34~kg$
$1.34~kg$ of water was in the vessel.
