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Step-by-Step Solution
Step 1: Identify the Problem
We want to find the time required for an electric kettle to heat 1 kg of water from 20°C to 100°C, given that the kettle operates at 220 V and draws 4 A current.
Step 2: List the Known Data
Current, I = 4 A
Voltage, V = 220 V
Mass of water, m = 1 kg
Initial temperature of water, $T_i = 20^\circ \text{C}$
Final temperature of water, $T_f = 100^\circ \text{C}$
Specific heat capacity of water, $c = 4200 \, \text{J/kg K}$ (or 1 cal/g·°C, which is approximately 4.2 J/g·°C)
Step 3: Calculate the Electric Power
Electric power, $P$, is given by the product of voltage and current:
$P = V \times I = 220 \times 4 = 880 \,\text{W}$
(Note: $1 \,\text{W} = 1 \,\text{J/s}$)
Step 4: Calculate the Heat Required
The heat needed to raise the temperature of water from $T_i$ to $T_f$ is given by:
$Q = m \times c \times \Delta T$
Here,
$\Delta T = T_f - T_i = 100 - 20 = 80^\circ \text{C}.$
In SI units:
$Q = 1 \,\text{kg} \times 4200 \,\text{J/kg K} \times 80 \,\text{K} = 336{,}000 \,\text{J}
Step 5: Determine the Time Required
Since power $P$ is the rate at which energy is supplied, the time taken, $t$, can be found using:
$t = \frac{Q}{P} = \frac{336{,}000 \,\text{J}}{880 \,\text{J/s}}
$t \approx 381.8 \,\text{s}
Step 6: Convert the Time into Minutes
To convert seconds into minutes, we divide by 60:
$t \approx \frac{381.8}{60} \approx 6.36 \,\text{minutes}
Rounding suitably, we get approximately 6.3 minutes.
Final Answer
The time taken to boil 1 kg of water from 20°C to 100°C is about 6.3 minutes.
