Question

An electric kettle is filled with 1.3 kg of water at 20°C. The power of the heating coil of the kettle is 2.0 kW. After switching it on the water begins to boil in 220s. If the kettle was kept on for a further interval of $\Delta t$ it was observed that only 200g water remained in the kettle and remaining water vaporized (the vapour is allowed to escape through a small vent). The specific latent heat of vaporization of water at 100°C (boiling point) is $L = 2.26$ kJ/g. Calculate the specific heat capacity of water and the interval $\Delta t$. Assume that heat supplied by the heater is completely absorbed by the water.

• A. Specific heat capacity of water = $\frac{55000}{13}$ J/kg°C, $\Delta t$ = 1243 s
• B. Specific heat capacity of water = $\frac{55000}{13}$ J/kg°C, $\Delta t$ = 220 s
• C. Specific heat capacity of water = 4200 J/kg°C, $\Delta t$ = 1243 s
• D. Specific heat capacity of water = 4200 J/kg°C, $\Delta t$ = 220 s

Answer 1

Answer: (A)

Specific heat capacity of water = $\frac{55000}{13}$ J/kg°C, $\Delta t$ = 1243 s

Answer 2

Step 1: Calculate the specific heat capacity of water ($c_w$)

The heat supplied by the kettle in the first 220 seconds is used to raise the temperature of 1.3 kg of water from 20°C to 100°C. Power of the kettle, $P = 2.0 \, \text{kW} = 2000 \, \text{W}$. Time, $t_1 = 220 \, \text{s}$. Heat supplied, $Q_1 = P \times t_1 = 2000 \, \text{W} \times 220 \, \text{s} = 440000 \, \text{J}$.

The heat required to raise the temperature of water is given by: $Q_1 = m \times c_w \times \Delta T$ where $m = 1.3 \, \text{kg}$, and $\Delta T = 100^\circ\text{C} - 20^\circ\text{C} = 80^\circ\text{C}$.

So, $440000 \, \text{J} = 1.3 \, \text{kg} \times c_w \times 80^\circ\text{C}$. $c_w = \frac{440000}{1.3 \times 80} \, \text{J/kg}^\circ\text{C} = \frac{440000}{104} \, \text{J/kg}^\circ\text{C} = \frac{55000}{13} \, \text{J/kg}^\circ\text{C}$. Approximately, $c_w \approx 4230.77 \, \text{J/kg}^\circ\text{C}$.

Step 2: Calculate the interval $\Delta t$

After the water reaches boiling point, the kettle is kept on for an additional time $\Delta t$. During this time, a portion of the water vaporizes. Initial mass of water = 1.3 kg. Remaining mass of water = 200 g = 0.2 kg. Mass of water vaporized, $m_{vap} = 1.3 \, \text{kg} - 0.2 \, \text{kg} = 1.1 \, \text{kg}$.

The specific latent heat of vaporization of water is given as $L = 2.26 \, \text{kJ/g}$. Convert $L$ to J/kg: $L = 2.26 \, \text{kJ/g} = 2.26 \times 10^3 \, \text{J/g} = 2.26 \times 10^6 \, \text{J/kg}$.

The heat required to vaporize 1.1 kg of water is: $Q_{vap} = m_{vap} \times L = 1.1 \, \text{kg} \times 2.26 \times 10^6 \, \text{J/kg} = 2.486 \times 10^6 \, \text{J}$.

This heat is supplied by the kettle during the interval $\Delta t$. Heat supplied, $Q_2 = P \times \Delta t = 2000 \, \text{W} \times \Delta t$.

Equating $Q_2$ and $Q_{vap}$: $2000 \, \text{W} \times \Delta t = 2.486 \times 10^6 \, \text{J}$. $\Delta t = \frac{2.486 \times 10^6}{2000} \, \text{s} = \frac{2486000}{2000} \, \text{s} = 1243 \, \text{s}$.

Therefore, the specific heat capacity of water is $\frac{55000}{13} \, \text{J/kg}^\circ\text{C}$ and the interval $\Delta t$ is 1243 s.