A 45kg sample of water absorbs 364kJ of heat. If the water was initially at 28.2 Celcius, what is its final temperature? How do I figure this out....HELP!
Note the correct spelling of celsius.
364,000 = mass H2O x specific heat H2O x (Tfinal-Tinitial). Substitute and solve for Tf.
To find the final temperature of the water, you can use the equation:
Q = mcΔT
Where:
Q is the heat absorbed by the water (364 kJ)
m is the mass of water (45 kg)
c is the specific heat capacity of water (4.186 J/g°C, or 4.186 kJ/kg°C)
ΔT is the change in temperature (final temperature - initial temperature)
First, convert the heat absorbed from kJ to J:
364 kJ × 1000 J/kJ = 364,000 J
Next, convert the specific heat capacity from kJ/kg°C to J/kg°C:
4.186 kJ/kg°C × 1000 J/kJ = 4,186 J/kg°C
Now, rearrange the equation to solve for ΔT:
Q = mcΔT
ΔT = Q / (mc)
Plug in the values:
ΔT = 364,000 J / (45 kg × 4,186 J/kg°C)
Calculate:
ΔT = 1.94°C
Finally, add the change in temperature to the initial temperature to find the final temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 28.2°C + 1.94°C
Therefore, the final temperature of the water is approximately 30.14°C.