how many kcals are required to cool one gallon/liter of water from 32degrees celcius to 7 degrees celcius?
To calculate the number of kilocalories (kcal) required to cool one gallon/liter of water from 32 degrees Celsius to 7 degrees Celsius, you need to know the specific heat capacity of water and the mass of the water you are trying to cool.
The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius (J/g°C), which is equivalent to 1 kilocalorie per kilogram per degree Celsius (kcal/kg°C).
First, you need to convert the volume of water from gallons/liters to kilograms. The density of water is about 1 kilogram per liter, so one gallon (3.785 liters) is approximately equal to 3.785 kilograms.
Next, calculate the mass of the water in grams by multiplying the mass in kilograms by 1000, as there are 1000 grams in a kilogram. In this case, 3.785 kilograms would be equal to 3785 grams.
Now, you can use the formula:
Q = m * c * ΔT
where:
Q is the energy transferred (kcal)
m is the mass of the water (g)
c is the specific heat capacity of water (kcal/kg°C)
ΔT is the change in temperature (°C)
Plugging in the values:
Q = 3785 g * 1 kcal/kg°C * (32°C - 7°C)
Q = 3785 g * 1 kcal/kg°C * 25°C
Q = 94625 kcal
Therefore, it would require approximately 94,625 kilocalories to cool one gallon/liter of water from 32 degrees Celsius to 7 degrees Celsius.