A 249-gram sample of water is heated from 2° C to 85° C. Calculate the amount
of heat absorbed (in kJ) by the water.
Please explain.
mass x specific heat water x delta T.
To calculate the amount of heat absorbed by the water, we can use the equation:
Q = m * c * ΔT
Where:
Q is the amount of heat absorbed,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
Now, let's substitute the given values into the equation:
- The mass of the water, m, is given as 249 grams.
- The change in temperature, ΔT, is the final temperature minus the initial temperature. So, ΔT = 85°C - 2°C = 83°C.
The specific heat capacity of water, c, is approximately 4.18 J/g°C. However, we need to convert the mass from grams to kilograms to match the units of the specific heat capacity.
To convert grams to kilograms, divide the mass by 1000:
m = 249 g ÷ 1000 = 0.249 kg
Now, let's calculate the amount of heat absorbed:
Q = 0.249 kg * 4.18 J/g°C * 83°C = 86.8222 J
Since the question asks for the amount of heat absorbed in kilojoules (kJ), we need to convert the energy from joules to kilojoules by dividing it by 1000:
Q = 86.8222 J ÷ 1000 = 0.0868 kJ
Therefore, the amount of heat absorbed by the water is approximately 0.0868 kJ.