Question

600 ml of water in an electric kettle is heated from 20.0°c to 85.0°c to make a cup of tea. how much thermal energy is absorbed by the water

Answer 1

the thermal energy absorbed by the water is approximately 155.22 kJ.

Step-by-step explanation:

The thermal energy absorbed by the water can be calculated using the formula:

Q = mcΔT

where:

Q is the thermal energy absorbed by the water (in Joules)

m is the mass of the water (in kilograms)

c is the specific heat capacity of water (in Joules per kilogram per degree Celsius)

ΔT is the change in temperature of the water (in degrees Celsius)

We know that the mass of the water is 600 grams or 0.6 kg, the specific heat capacity of water is 4.18 J/g°C, and the change in temperature is (85.0 - 20.0) = 65.0°C.

Substituting these values into the formula, we get:

Q = (0.6 kg) x (4.18 J/g°C) x (65.0°C)

Q = 155.22 kJ

Answer 2

the thermal energy absorbed by the water is 163,704 Joules.

Step-by-step explanation:

The thermal energy absorbed by the water can be calculated using the formula:

Q = m * c * ΔT

where Q is the thermal energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

In this case, the mass of the water is 600 grams (since 1 mL of water has a mass of 1 gram), the specific heat capacity of water is 4.184 J/g°C, and the change in temperature is 85.0°C - 20.0°C = 65.0°C.

So, plugging in the values, we get:

Q = 600 g * 4.184 J/g°C * 65.0°C

Q = 163,704 J

Therefore, the thermal energy absorbed by the water is 163,704 Joules.