Final answer:
To calculate the change in energy for cooling 75.0 grams of water from 31.0°C to 21.6°C, the formula Q = mcΔT is used, resulting in approximately -2955 J, which does not match any of the provided options.
Step-by-step explanation:
To calculate the change in energy when 75.0 grams of water drops from 31.0°C to 21.6°C, we will use the formula for heat transfer Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature.
First, we need to calculate the change in temperature (ΔT):
ΔT = 31.0°C - 21.6°C = 9.4°C
Next, we use the formula to find Q:
Q = (75.0 g) * (4.184 J/g°C) * (9.4°C)
Q = (75.0 g) * (4.184 J/g°C) * (9.4°C)
Q = 2955.24 J
However, since this is a decrease in temperature, the energy change will be negative:
Q = -2955.24 J
After converting to kilojoules (by dividing by 1000), we get:
Q = -2.95524 kJ ≈ -2955 J
None of the options given correctly represents the change in energy. The correct energy change is approximately -2955 J.