Using the values given, calculate the amount of heat that is liberated when a 2.7-kg bucket of steam at 100.0°C is cooled to give a 2.7-kg bucket of water at 28.0°C. (Condensing steam has −ÄvapH = −2440 J/g and cooling water has cliq = 4.184 J/g·°C.)

To calculate the amount of heat liberated when the steam is cooled to water, we need to calculate the heat lost in two steps:

Step 1: Heat liberated during condensation of steam to water.
Step 2: Heat lost during cooling of water from 100.0°C to 28.0°C.

Let's calculate each step separately.

Step 1: Heat liberated during condensation of steam to water.
The given value for ΔvapH is −2440 J/g. However, we need to convert the mass from kg to g. So, multiply the mass of the steam (2.7 kg) by 1000 to get the mass in grams.
Mass of steam = 2.7 kg x 1000 g/kg = 2700 g

Now, calculate the heat liberated during condensation:
Heat liberated = mass of steam x ΔvapH
Heat liberated = 2700 g x (-2440 J/g)

Step 2: Heat lost during cooling of water from 100.0°C to 28.0°C.
The specific heat capacity of water (cliq) is 4.184 J/g°C. Again, convert the mass from kg to g.
Mass of water = 2.7 kg x 1000 g/kg = 2700 g

Now, calculate the heat lost during cooling:
Heat lost = mass of water x specific heat capacity x temperature change
Heat lost = 2700 g x 4.184 J/g°C x (100.0°C - 28.0°C)

Finally, add the heat liberated in Step 1 and the heat lost in Step 2 to get the total amount of heat liberated:
Total heat liberated = Heat liberated in Step 1 + Heat lost in Step 2

Now, just substitute the values and calculate the final answer.