What is the enthalpy change when 24.0 g of water is cooled from 40.0°C to 5.70°C? _________kJIs the process endothermic or exothermic?
To determine the enthalpy change when cooling water from one temperature to another, we need to use the equation:
ΔH = mcΔT
Where:
ΔH is the enthalpy change (in joules or 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, we need to determine the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Given:
m (mass of water) = 24.0 g
c (specific heat capacity of water) = 4.18 J/g°C
ΔT (change in temperature) = 5.70°C - 40.0°C
Now, let's substitute the given values into the equation:
ΔH = (24.0 g) x (4.18 J/g°C) x (-34.3°C)
Calculating this expression will give us the enthalpy change in joules. To convert it to kilojoules, divide the result by 1000.
Therefore, the enthalpy change is: ΔH = -2796.456 J = -2.797 kJ.
Since the value of ΔH is negative, it indicates an exothermic process.