A 45kg sample of water absorbs 349kJ of heat. If the water was initially at 25.5 degrees Celsius, what is its final temperature?
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
I would convert 349 kJ to 349,000 J and 45 kg H2O to 45,000 g and use 4.184 J/g*C for specific heat. Solve for Tf.
To find the final temperature of the water, we can use the equation:
Q = mcΔT
Where:
Q = amount of heat absorbed (349 kJ)
m = mass of water (45 kg)
c = specific heat capacity of water (4.18 kJ/kg°C)
ΔT = change in temperature (final temperature - initial temperature)
First, let's solve for ΔT:
Q = mcΔT
349 kJ = (45 kg)(4.18 kJ/kg°C)ΔT
Simplifying:
349 kJ = 187.1 kg°C ΔT
ΔT = 349 kJ / 187.1 kg°C
ΔT ≈ 1.866°C
Now, let's find the final temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 25.5°C + 1.866°C
Final temperature ≈ 27.366°C
Therefore, the final temperature of the water is approximately 27.366°C.