By how much would 1200 W kettle, which is 72% efficient raise the temperature of 0.60 kg of water in 3.0 min?

To determine the change in temperature of the water, we can use the formula:

Q = mc∆T

Where:
Q = heat energy (in joules)
m = mass of water (in kilograms)
c = specific heat capacity of water (approximately 4186 J/kg·°C)
∆T = change in temperature (in °C)

The heat energy (Q) can be calculated using the power (P) and time (t) through the equation:

Q = Pt

The power of the kettle is given as 1200 W, and the time is given as 3.0 minutes. Firstly, we need to convert the time from minutes to seconds:

t = 3.0 minutes = 3.0 × 60 seconds = 180 seconds

Now, let's calculate the heat energy:

Q = Pt = 1200 W × 180 s = 216,000 J

Since the kettle is 72% efficient, only 72% of the heat energy is transferred to the water. Thus, the actual heat energy transferred to the water is:

Q_actual = (72/100) × Q = 0.72 × 216,000 J = 155,520 J

Next, we can use the formula Q = mc∆T to find the change in temperature (∆T):

∆T = Q_actual / (mc)

The mass of water (m) is given as 0.60 kg. Plugging in the values, we get:

∆T = 155,520 J / (0.60 kg × 4186 J/kg·°C) ≈ 62.23 °C

Therefore, the kettle, with 72% efficiency, would raise the temperature of 0.60 kg of water by approximately 62.23 °C in 3.0 minutes.