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
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.
To calculate the heat required to heat a certain amount of water, you can use the following formula:
Q = m * C * ΔT
Where:
Q = heat required (in kilojoules)
m = mass of water (in kilograms)
C = specific heat capacity of water (4.18 J/g°C or 4.18 kJ/kg°C)
ΔT = temperature change (in °C)
First, let's convert the given temperature values to Celsius:
Initial temperature = (85°F - 32) * 5/9 = 29.4°C
Final temperature = (212°F - 32) * 5/9 = 100°C
Next, we need to determine the mass of the water. We know the volume is 56 L, but we need the mass. The density of water is approximately 1 gram/mL, so:
Mass of water = volume of water * density of water
= 56 L * 1000 g/L
= 56,000 grams
= 56 kg
Now we can calculate the heat required:
Q = m * C * ΔT
= 56 kg * 4.18 kJ/kg°C * (100°C - 29.4°C)
= 56 kg * 4.18 kJ/kg°C * 70.6°C
≈ 16533.12 kJ
Therefore, approximately 16533.12 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.