To calculate the time required to boil 1 kg of water using an electric kettle, we need to use the formula for electrical energy and the specific heat capacity of water.
Given data:
• Current I=4 AI = 4 \, \text{A}
• Voltage V=220 VV = 220 \, \text{V}
• Mass of water m=1 kgm = 1 \, \text{kg}
• Initial temperature of water Tinitial=20∘ CT_{\text{initial}} = 20^\circ \, \text{C}
• Final temperature of water Tfinal=100∘ CT_{\text{final}} = 100^\circ \, \text{C}
• Specific heat capacity of water c=4200 J/kg∘Cc = 4200 \, \text{J/kg}^\circ \text{C}
• The temperature difference ΔT=Tfinal−Tinitial=100∘ C−20∘ C=80∘ C\Delta T = T_{\text{final}} - T_{\text{initial}} = 100^\circ \, \text{C} - 20^\circ \, \text{C} = 80^\circ \, \text{C}
Step 1: Calculate the energy required to heat the water.
The heat energy QQ required to raise the temperature of the water can be calculated using the formula:
Q=mcΔTQ = mc\Delta T
Substitute the values:
Q=1 kg×4200 J/kg∘C×80∘ CQ = 1 \, \text{kg} \times 4200 \, \text{J/kg}^\circ \text{C} \times 80^\circ \, \text{C} Q=336,000 JQ = 336,000 \, \text{J}
Step 2: Calculate the power supplied by the kettle.
The power PP supplied by the electric kettle can be calculated using the formula:
P=IVP = IV
Substitute the values:
P=4 A×220 VP = 4 \, \text{A} \times 220 \, \text{V} P=880 W=880 J/sP = 880 \, \text{W} = 880 \, \text{J/s}
Step 3: Calculate the time required to heat the water.
Now, the time tt required to boil the water is given by:
t=QPt = \frac{Q}{P}
Substitute the values:
t=336,000 J880 J/st = \frac{336,000 \, \text{J}}{880 \, \text{J/s}} t≈381.8 secondst \approx 381.8 \, \text{seconds}
Step 4: Convert the time to minutes.
t=381.8 seconds60t = \frac{381.8 \, \text{seconds}}{60} t≈6.36 minutest \approx 6.36 \, \text{minutes}
Final Answer:
The time required to boil 1 kg of water is approximately 6.4 minutes.
So, the correct answer is: D) 6.4 min.