When the power supplying a 30.0 gallon water heater was turned off, the temperature dropped from 75.0 degrees celsius to 22.5 degrees. How much heat was lost to the surrounding?
I know I need to find the energy and or heat lost, but how will i get the mass of H2O used to complete the equation?
Don't we assume that if it is a 30 gallon water heater that it will be full of H2O so it will hold 30 gallons. 1 gallon = 3.785 L.
I could multiply 30 by 3.785 L to find the volume, but to find the mass of water used I also need the density of the water which is not given. Is there a different equation I should be using other than...
qH2O = mH2O x CsH2O x delta T ?
No. Density is water generally is 1.0 g/mL for most of these problems. Therefore, volume in mL = mass in grams.
Okay, thank you for clarifying. :)
To find the heat lost, you need to calculate the amount of energy required to change the temperature of the water from 75.0 degrees Celsius to 22.5 degrees Celsius.
The formula to calculate the energy is:
Q = mcΔT
Where:
Q is the heat energy (in Joules),
m is the mass of the water (in kilograms),
c is the specific heat capacity of water (approximately 4.18 J/g°C), and
ΔT is the change in temperature (in degrees Celsius).
First, convert the volume of the water heater from gallons to liters since the specific heat capacity of water is given in grams.
1 gallon is approximately equal to 3.78541 liters.
So, 30.0 gallons is equal to 30.0 x 3.78541 liters = 113.5623 liters.
Next, find the mass of the water using the density of water, which is 1 gram per milliliter.
1 liter of water is equal to 1000 milliliters, and 1 milliliter of water is equal to 1 gram.
So, the mass of the water is 113.5623 liters x 1000 grams/liter = 113562.3 grams.
Now, convert the mass of the water from grams to kilograms by dividing by 1000.
113562.3 grams / 1000 = 113.5623 kilograms.
Finally, substitute the values into the formula:
Q = (113.5623 kg) x (4.18 J/g°C) x (75.0°C - 22.5°C)
Now, calculate the result to find how much heat was lost to the surroundings.