To solve this problem, we need to use the formula for heat transfer:
Q = mcΔT
Where:
Q is the heat transfer (in Joules)
m is the mass of water (in kilograms)
c is the specific heat capacity of water (in J/kg·°C)
ΔT is the change in temperature (in °C)
We can rearrange the formula to solve for time:
t = Q / P
Where:
t is the time (in seconds)
Q is the heat transfer (in Joules)
P is the power of the kettle (in Watts)
Given:
P = 2000 W
m = 2.0 kg
ΔT = (100 - 20) °C = 80 °C (200°C - 20°C)
First, let's calculate the heat transfer:
Q = mcΔT
Q = 2.0 kg * 4186 J/kg·°C * 80 °C
Q = 669,760 J
Now, let's calculate the time:
t = Q / P
t = 669,760 J / 2000 W
t ≈ 334.88 seconds
Therefore, it will take approximately 334.88 seconds for the temperature of the water to rise from 20°C to 100°C.
An electric kettle with negligible heat capacity is
rated at 2000 W. If 2.0kg of water is put in it,
how long will it take the temperature of water to
rise from 200C to 1000C?
1 answer
1 answer