Solution
Let $P$ be the constant power of the heater and $P_{diss}$ be the constant rate of heat dissipation to the surroundings.
Phase 1: Heating (Heater On)
Time $\Delta t_1 = 1.0 \text{ min} = 60 \text{ s}$. Temp rise $\Delta T = 65 – 60 = 5^\circ\text{C}$.
Energy Balance: Heat Supplied – Heat Dissipated = Heat used to raise temp
$$ (P – P_{diss}) \Delta t_1 = ms \Delta T $$
$$ P – P_{diss} = \frac{ms(5)}{60} = \frac{ms}{12} \quad \dots(1) $$
Phase 2: Cooling (Heater Off)
Time $\Delta t_2 = 9.0 \text{ min} = 540 \text{ s}$. Temp fall $\Delta T = 65 – 60 = 5^\circ\text{C}$.
Energy Balance: Heat Dissipated = Heat released by water
$$ P_{diss} \Delta t_2 = ms \Delta T $$
$$ P_{diss} = \frac{ms(5)}{540} = \frac{ms}{108} \quad \dots(2) $$
Finding Proportion:
We need the ratio of Heat Dissipated to Heat Received during heating.
This is effectively the ratio of powers $P_{diss} / P$.
Substitute (2) into (1):
$$ P – \frac{ms}{108} = \frac{ms}{12} $$
$$ P = \frac{ms}{12} + \frac{ms}{108} = \frac{9ms + ms}{108} = \frac{10ms}{108} $$
Ratio $\frac{P_{diss}}{P}$:
$$ \frac{ms/108}{10ms/108} = \frac{1}{10} $$
Percentage = $\frac{1}{10} \times 100\% = 10\%$
Correct Option: (b) 10%