How many kilojoules of heat are absorbed when 1.00L of water is heated from 18 degrees C to 85 degrees C?
To calculate the amount of heat absorbed by heating water, we need to use the formula:
Q = m × c × ΔT
Where:
Q is the heat absorbed (in kilojoules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in joules per gram per degree Celsius),
ΔT is the change in temperature (in degrees Celsius).
First, let's calculate the mass of water. The density of water is approximately 1 gram per milliliter (or 1 kilogram per liter). Therefore, the mass of 1.00 liter of water is 1000 grams (or 1 kilogram).
Next, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 joules per gram per degree Celsius (or 4.18 kilojoules per kilogram per degree Celsius).
Now, we can substitute the values into the formula:
Q = m × c × ΔT
Q = 1000 g × 4.18 J/g°C × (85°C - 18°C)
Note that we convert the units of mass to grams and the specific heat capacity to joules per gram per degree Celsius.
Q = 1000 g × 4.18 J/g°C × 67°C
Finally, let's calculate the value:
Q ≈ 281,260 J
To convert the result to kilojoules, divide by 1000:
Q ≈ 281.26 kJ
Therefore, approximately 281.26 kilojoules of heat are absorbed when 1.00 liter of water is heated from 18 degrees Celsius to 85 degrees Celsius.