What is the final temperature (degrees celcius) of 1.36g of water with an initial temperature of 23.0degrees celcius after 6.420J of heat is added to it?
m=dv
mol C2H2=m/molar mass
heat(q)= (delta H)(mol)
q = [mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0
To find the final temperature of the water, we can use the equation:
q = m * c * ΔT
Where:
q is the heat energy (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of water (which is approximately 4.18 J/g·°C),
ΔT is the change in temperature (final temperature - initial temperature).
First, let's calculate the ΔT using the equation:
ΔT = q / (m * c)
Since the initial temperature is 23.0°C, we have:
ΔT = q / (m * c) = 6.420 J / (1.36 g * 4.18 J/g·°C)
Now, let's evaluate this to find the change in temperature:
ΔT ≈ 1.413 °C
Finally, to find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT = 23.0 °C + 1.413 °C
Therefore, the final temperature of 1.36g of water, after 6.420J of heat is added to it, is approximately 24.413 °C.