A 45-kg sample of water absorbs 345 kj of heat. If the water was initially at 22.1 degrees c what is its final temperature
To determine the final temperature of the water, we need to use the equation:
q = mcΔT
where:
q is the heat absorbed by the water,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
Rearranging the equation, we can solve for ΔT:
ΔT = q / (mc)
First, let's calculate the specific heat capacity of water. The specific heat capacity of water is approximately 4.184 J/g°C.
c = 4.184 J/g°C
Since the mass of the water is given in kilograms, we need to convert it to grams:
m = 45 kg * 1000 = 45000 g
Now we can substitute the values into the equation:
ΔT = (345000 J) / (45000 g * 4.184 J/g°C)
Simplifying the equation:
ΔT = 1.617 °C
To find the final temperature, we can add ΔT to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 22.1°C + 1.617°C
Final temperature ≈ 23.717°C
Therefore, the final temperature of the water is approximately 23.717°C.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Tfinal is the only uniknown.