Answer
(a) The oven puts energy into the water at a rate of 797 J/s
(b) 5.3 grams of water will boil away.
Work Step by Step
The energy $Q$ required to raise the temperature of a substance is:
$Q = mc~\Delta T$
We can find the heat energy gained by the water.
$Q = m_w~c_w~\Delta T$
$Q = (0.250~kg)(4186~J/kg~C^{\circ})(100~^{\circ}C-20~^{\circ}C)$
$Q = 83,720~J$
We can find the rate at which the oven puts energy into the water.
$\frac{Q}{t} = \frac{83,720~J}{105~s}$
$\frac{Q}{t} = 797~J/s$
The oven puts energy into the water at a rate of 797 J/s.
(b) We can find the additional energy provided to the water if the oven runs for an additional 15 seconds.
$Q = (797~J/s)(15~s) = 11,955~J$
We can find the mass of water which boils away.
$Q = m~L$
$m = \frac{Q}{L}$
$m = \frac{11,955~J}{2.26\times 10^6~J/kg}$
$m = 0.0053~kg = 5.3~g$
5.3 grams of water will boil away.