What is the enthalpy change when 19.0 g of water is cooled from 33.0°C to 2.70°C?
To calculate the enthalpy change when water is cooled from one temperature to another, we can use the equation:
ΔH = mcΔT
Where:
ΔH is the enthalpy change (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C),
ΔT is the change in temperature (in °C).
In this case, we need to find the enthalpy change (ΔH) when 19.0g of water is cooled from 33.0°C to 2.70°C.
First, let's calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 2.70°C - 33.0°C
ΔT = -30.3°C
Next, we need to determine the specific heat capacity of water (c). The specific heat capacity of water is 4.18 J/g°C.
Now, we can plug in the values into the formula to calculate the enthalpy change:
ΔH = (mass)(specific heat capacity)(ΔT)
ΔH = (19.0g)(4.18 J/g°C)(-30.3°C)
Calculating the enthalpy change:
ΔH = -2,285.422 J (since the value of ΔH is negative, it indicates that heat is being released or lost)
Therefore, the enthalpy change when 19.0 g of water is cooled from 33.0°C to 2.70°C is approximately -2,285.422 J.