Answer
$67.08$
Work Step by Step
Using the equation $q=mc\Delta T$, we can rearrange the equation and solve for the change in temperature ; $\Delta T = \frac{q}{mc}$. The specific heat capacity of water is $4.184 \frac{J}{g\times^{\circ}C}$ and the mass of the water in this case is 29.5 kg or 29500 grams since we are told we have 29.5 L and since the density of water is 1 g/L we can use the simple relationship to get 29500 grams. Also we have to convert the heat in terms of Joules by using the following relationship:
$1 kWh = 3.6\times10^{6} J$, therefore, $2.3 kWh = 8.28\times10^{6} J$,
$\Delta T = \frac{8.28\times10^{6} J }{29500g\times4.184}$
$\Delta T = 67.08$
Therefore, the change in temperature is $67.08$
