Answer
The actual drop in heat output would smaller than that calculated in (a).
Work Step by Step
We can find an expression for the original heat output:
$P_0 = \frac{(115~V)^2}{R}$
We can find an expression for the new heat output:
$P_f = \frac{(110~V)^2}{R}$
We can find the ratio of $\frac{P_f}{P_0}$:
$\frac{P_f}{P_0} = \frac{\frac{(110~V)^2}{R}}{\frac{(115~V)^2}{R}} = \frac{(110~V)^2}{(115~V)^2} = 0.915$
The heat output decreases by 8.5%
However, if the heat output decreases, then the resistance would operate at a lower temperature. At a lower temperature, the resistance decreases.
Then the actual value of $P_f$ would be higher than $\frac{(110~V)^2}{R}$
In this case, the ratio $\frac{P_f}{P_0}$ would be higher than the $0.915$ that was calculated above.
Therefore, the actual drop in heat output would smaller than that calculated in (a).
