If an aquarium tank holds 100 litres of water, calculate how much heat energy would be required to heat the tank from 18.2 Degrees(C) to 32 Degrees(C)
I cant for the life of me figure out how to complete this question as our teacher hasn't properly gone over this section.
If you assume the density of water is 1.00 g/mL, then 1000 L H2O will have a mass of 1000 grams.
Then q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
To clarify, 100 L weigh 100 kg or 1x10^5 g
Yes and I'm pleased that you caught that (those) typos.
100 L is 100E3 mL and
100E3 mL x 1.00 g/mL = 1E5 g.
To calculate the heat energy required to heat the tank from one temperature to another, we need to use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy (in joules)
m is the mass of the substance (in kilograms)
c is the specific heat capacity (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
In this case, we need to calculate the heat energy required to heat the 100 liters of water from 18.2 degrees Celsius to 32 degrees Celsius.
Step 1: Convert the volume of water from liters to kilograms.
Density of water = 1 kg/L
So, 100 liters of water would be equal to 100 kilograms.
Step 2: Determine the specific heat capacity of water.
The specific heat capacity of water is approximately 4.186 joules per gram per degree Celsius, or 4,186 joules per kilogram per degree Celsius.
Step 3: Calculate the change in temperature.
ΔT = (Final temperature) - (Initial temperature)
ΔT = 32°C - 18.2°C
ΔT = 13.8°C
Step 4: Plug the values into the formula and calculate the heat energy.
Q = m * c * ΔT
Q = 100 kg * 4,186 J/(kg°C) * 13.8°C
Q ≈ 5,781,480 joules
Therefore, approximately 5,781,480 joules of heat energy would be required to heat the tank from 18.2 degrees Celsius to 32 degrees Celsius.