Final answer:
It will take approximately 59.28 seconds for a 3.50 kW electric heater to raise the temperature of 0.620 kg of water from 20.0°C to 100.0°C, assuming no heat loss.
Step-by-step explanation:
To calculate how long it will take to raise the temperature of 0.620 kg of water from 20.0°C to 100.0°C using a 3.50 kW electric heater, we use the formula Q = mcΔT, where 'Q' is the heat energy required, 'm' is the mass of the water, 'c' is the specific heat capacity of water, and ΔT is the change in temperature. For water, 'c' is typically 4.184 J/g°C (or 4184 J/kg°C). Firstly, convert the mass to grams (since the specific heat capacity is per gram), which gives 620g (0.620 kg × 1000 g/kg).
Now calculate the energy needed:
Q = (620 g)(4.184 J/g°C)(100.0°C - 20.0°C) = (620)(4.184)(80) = 207,488 J
To find the time 't' it takes, use the power 'P' of the kettle, which is the energy per unit time:
t = Q / P = 207,488 J / 3,500 J/s = 59.28 s
Thus, it will take approximately 59.28 seconds to heat the water to the desired temperature. Remember that this calculation assumes no heat losses to the environment, which is not the case in actual scenarios.