A lukewarm bottle containing 800 grams of water is placed in a refrigerator. How many calories of heat must be removed from the water to chill its temperature from 55 C to 4 C?

To calculate the amount of heat that needs to be removed from the water, we need to use the specific heat capacity formula.

The specific heat capacity of water is approximately 4.18 J/g°C. This means that it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

First, we need to calculate the temperature change:
ΔT = Tfinal - Tinitial = 4°C - 55°C = -51°C

Next, we need to calculate the heat energy:
Q = m * c * ΔT

Where:
Q is the heat energy in calories,
m is the mass of the water in grams,
c is the specific heat capacity of water, and
ΔT is the temperature change in Celsius.

Substituting the values:
Q = 800 grams * 4.18 J/g°C * (-51°C)

Converting J to calories by dividing by 4.184:
Q = (800 g * 4.18 J/g°C * (-51°C)) / 4.184

Calculating the value:
Q ≈ -39937.733 calories

Since the energy value is negative, it means heat must be removed from the water. Thus, approximately 39937.733 calories of heat must be removed from the water to chill its temperature from 55°C to 4°C.