If a 250 mL sample of water was cooled from 50.0˚C to 25˚C, how much heat was lost (in joules)?
q = mass H2O x specific heat H2O x delta T.
To calculate the heat lost by the water sample, you can use the equation:
Q = m * C * ΔT,
where Q represents the heat lost (in joules), m is the mass of the water (in grams), C is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (in °C).
First, convert the volume of the water sample to grams using its density. The density of water is approximately 1 g/mL. Therefore, the mass of the water sample is:
m = volume * density = 250 mL * 1 g/mL = 250 grams.
Next, calculate the change in temperature:
ΔT = final temperature - initial temperature = 25°C - 50°C = -25°C.
Since the temperature decreased, you have a negative value for ΔT.
Now, you can substitute the values into the equation:
Q = 250 g * 4.18 J/g°C * (-25°C).
Calculating this, you get:
Q = -26,125 J.
Therefore, the amount of heat lost by the water sample is approximately -26,125 joules. Note that the negative sign indicates the heat has been lost.