## Essential University Physics: Volume 1 (3rd Edition)

We first must find how much energy is needed to bring the water to a boil: $Q = mc\Delta t = (2.5)(4184)(90)=941,400\ J \approx 941 \ kJ$ Recall, the energy supplied is equal to the integral of P with respect to time. Note, we must substitute a "dummy variable" in for t, for the same variable cannot be in both the expression and in the bounds of the integral. Thus, we find: $941 = \int_0^t 1.1 + 2.3 s$ $0 = 1.15t^2 + 1.1t -941$ Using either a calculator or the quadratic formula, we obtain: $t = 28 \ min$