What is the enthalpy change when 19.0 g of water is cooled from 33.0°C to 2.70°C?
q = mass x specific heat x (Tfinal-Tniitial)
q= 2409.88
is that right?
I get 2408.73 J but we aren't allowed but three significant figures and I would round that to 2.41 kJ.
To calculate the enthalpy change when water is cooled, you need to use the formula:
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)
To begin, you need to determine the mass (m) of water. In this case, it is given as 19.0 grams.
Next, you need to find the specific heat capacity (c) of water. The specific heat capacity of water is approximately 4.18 J/g°C. This value represents the amount of heat energy required to raise the temperature of one gram of water by one degree Celsius.
Finally, you need to calculate the change in temperature (ΔT). In this case, the water is cooled from 33.0°C to 2.70°C. Therefore, the change in temperature is:
ΔT = Final temperature - Initial temperature
ΔT = 2.70°C - 33.0°C
Now, you can substitute the values into the formula to calculate the enthalpy change (ΔH):
ΔH = (19.0 g) * (4.18 J/g°C) * (2.70°C - 33.0°C)
Calculating this expression will give you the enthalpy change in Joules.