How many kilojoules of heat are absorbed when 1.2 L of water is heated from 24oC to 86oC?
Heat, Q
=mcΔT
=1200*4.184*(86-24)°
=... joules
Divide by 1000 to convert to kj.
To calculate the amount of heat absorbed when raising the temperature of water, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat absorbed or released
m is the mass of the substance (in this case, water)
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, let's find the mass of water. We know the volume is 1.2 L, but we need to convert it to grams since specific heat capacity is usually given in J/g°C:
Density of water = 1 g/mL (approximately)
Mass = Volume * Density
So, the mass of water = 1.2 L * 1000 mL/L * 1 g/mL = 1200 grams
Next, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Now we can substitute the values into the formula:
Q = m * c * ΔT
Q = 1200 g * 4.18 J/g°C * (86°C - 24°C)
Calculating this:
Q = 1200 g * 4.18 J/g°C * 62°C
Q = 301,464 J
To convert joules to kilojoules, divide by 1000:
301,464 J / 1000 = 301.464 kJ
Therefore, approximately 301.464 kilojoules of heat are absorbed when 1.2 L of water is heated from 24°C to 86°C.