If a 375 mL sample of water was cooled from 37.5°C to 0°C, how much heat was lost (in joules)
J = mass x specific heat x (Tfinal-Tinitial)
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To calculate the amount of heat lost by the water, you can use the heat capacity equation:
Q = m * c * ΔT
where:
Q is the quantity of heat transferred in joules (J),
m is the mass of the substance in grams (g),
c is the specific heat capacity of the substance in J/g°C, and
ΔT is the change in temperature in °C.
In this case, the mass of the water sample is given as 375 mL. We need to convert this volume to grams, assuming the density of water is 1 g/mL:
Mass of water = Volume of water * Density of water
Mass of water = 375 mL * 1 g/mL
Mass of water = 375 g
The specific heat capacity of water is approximately 4.18 J/g°C.
ΔT is the change in temperature, which is the final temperature minus the initial temperature:
ΔT = Final temperature - Initial temperature
ΔT = 0°C - 37.5°C
ΔT = -37.5°C
Now we can substitute these values into the equation:
Q = m * c * ΔT
Q = 375 g * 4.18 J/g°C * (-37.5°C)
Calculating this expression, we find that the amount of heat lost by the water is:
Q = -58,593.75 J
Therefore, the water lost approximately 58,593.75 joules of heat.