How much heat in kilojoules is required to heat 56 L of water from an initial temperature of 85°F to final temperature of 212°F

Answer 1

To calculate the amount of heat required to heat 56 L of water from an initial temperature of 85°F to a final temperature of 212°F, we can use the specific heat capacity of water.

The specific heat capacity of water is approximately 4.184 J/g°C or 4.184 kJ/kg°C.
The density of water is approximately 1 g/mL or 1000 kg/m³.

First, we need to convert the volume of water from liters to grams. Since the density of water is 1 g/mL, 56 L is equivalent to 56,000 grams.

Next, we calculate the temperature change by subtracting the initial temperature from the final temperature: 212°F - 85°F = 127°F.

Now we can calculate the amount of heat required using the formula:

Heat = mass * specific heat capacity * temperature change

Mass = volume * density = 56,000 g
Specific heat capacity = 4.184 kJ/kg°C
Temperature change = 127°F = 127°F * (5/9) = 69.4°C

Heat = 56,000 g * 4.184 kJ/kg°C * 69.4°C
Heat ≈ 16,326,848 kJ

Therefore, approximately 16,326,848 kilojoules (kJ) of heat is required to heat 56 L of water from an initial temperature of 85°F to a final temperature of 212°F.