Question

Paragraph for question nos. 17 to 19 By varying the voltage applied to the kettle, you can change power consumption P. Depending on the P, kettle with water can be heated to different maximum temperatures. This dependence is shown in Table-1:

Power P (in Watt)	0	100	200	300
Temperature T(in °C)	20	40	60	80

Table-2 shows the variation of the temperature with time when the kettle of power 300W is switched off. The heat capacity of the empty kettle $C_0$ = 100 J/K, specific heat of water $S_w$ = 4200 J/kg . K. The density of water $\rho$ = 1000 kg/m³. Table-2:

Time t (in sec)	0	60	300	600	1200	2400
Temperature T (in °C)	80	75	60	45	30	20

If the power consumption is 400W

• A. The water will come to the boiling point and start boiling
• B. The water will reach boiling point but will not boil
• C. The water will not reach boiling point but start boiling
• D. The water will neither reach boiling point nor start boiling.

Answer 1

Answer: (B)

The water will reach boiling point but will not boil

Answer 2

The rate of heat loss from the kettle to the surroundings can be modeled using Newton's Law of Cooling: $P_{loss} = k(T - T_{ambient})$.

From Table-2, the ambient temperature is $T_{ambient} = 20^\circ C$. Using data from Table-1, we can find the heat loss constant $k$: For P = 300W, T = 80°C: $300 = k(80 - 20) \implies 300 = 60k \implies k = 5 \text{ W/°C}$.

When the power consumption is 400W, the equilibrium temperature $T_{max}$ is reached when $P_{supplied} = P_{loss}$. $400 = 5(T_{max} - 20)$ $80 = T_{max} - 20$ $T_{max} = 100^\circ C$.

At $T = 100^\circ C$, the heat loss is $P_{loss} = 5(100 - 20) = 5(80) = 400 \text{ W}$. The net power supplied is $P_{net} = P_{supplied} - P_{loss} = 400 \text{ W} - 400 \text{ W} = 0 \text{ W}$. Since the net power is zero at the boiling point, there is no energy available for vaporization. Therefore, the water will reach the boiling point but will not boil.