Rather than do the math for you I will lay out how to do it and you can do the numbers yourself.
First, convert the volume of 25 gallons to milliliters:
Volume (gal) -> Volume (mL)
Second find the change in temperature:
T(high) - T(low) = dT (oC)
The calorie is defined as the amount of energy to raise one gram of water 1 degree Celsius.
Energy (calories) = Mass of Water (g) * dT (oC)
to find the mass of water convert volume to mass using the density equation D= m/V
Volume (mL) * Density (g/mL) = Mass of Water (g)
So the required energy is simply:
E (cal) =mass*dT
If you must report your answer in Joules (customary) you can convert calories to Joules using an established conversion constant.
E (cal) -> E (J)
46389000 j
The simple beginning is that the definition of a calorie is "the energy required to raise the temperature of 1g of water 1°C." Therefore, the energy required to raise 17g of water 32°C: 17*32=544 cal. However, the question asked about ice. There is an extra bit of energy required for the change of physical state. The energy required to convert 1 gram of ice at 0°C to liquid water at 0°C is called the "latent heat" and is equal to about 80 cal. To convert 17g of ice, we multiply this together: 17g * 80cal/g = 1360 cal. So, we add this energy required for the change of state to the energy required to raise the listed quantity to the required temperature and we get 544 cal + 1360 cal = 1904 cal, assuming no heat is lost to the environment. I hope this clarifies some things.
The amount of water whose temperature would change by 15 degrees Celsius when it absorbs 2646 joules of heat energy is 42,2g H2O.
the same
19.7 kJ
80cal/g
3.50 J
1935 J (apex)
E = m c (delta)T
15480.80
46389000 j
I believe it is Calorie.
1935 J (apex)
false its 1 degrees Celsius
21 Kg = 2100 grams to rise the temperature of this amount of water by 2 degrees Celsius you need 2*2100 = 4200 calories or 17572.8 Joules.
25degres celsius has more thermal energy
A normal bath full of water. Sorry but this depends on units of energy that you are familiar with-. I will use caldepending where you are taug The enery required to raise the temperature of the a 1kg of water in a kettle from zero degrees to 100 is 100 kilocalories The bath contains more at least 20 kg of water, and the energy required to raise the temperature of that volume of water by 55 degrees is 20 X 55 = 1100 kcals