You would have added approximately 27.86 gallons of 120 degree water to 17.14 gallons of 57 degree water. The math is as follows:
96=(120x+57y)/45 and x+y=45
y=45-x
96=(120x+57(45-x))/45
96=(120x+2565-57x)/45
96=(63x+2565)/45
96=1.4x+57
1.4x=39
x~27.86
y~45-27.86
y~17.14
Water is a solid at all temperatures from 0 degrees C downwards. That is at normal atmospheric pressure.
Depends on how pure the water is. If the water is 100% pure then then it will freeze at 0 degrees. IF it has a small amount of salt added to it, it might freeze about -3 degrees Depends on how pure the water is. If the water is 100% pure then then it will freeze at 0 degrees. IF it has a small amount of salt added to it, it might freeze about -3 degrees
In Fahrenheit the freezing temperature is 32 degrees. negative three degrees would be 35 degrees below freezing.
THE ANSWER IS 62.8 DEGREES.....
Above freezing water freezes at 0 degress celsius.
-3°F is 35 degrees below the freezing temperature of water, which is 32°F.
Which is most likely be the temperature of boiling water? 10 degress f 55 degrees f 78 degrees f 110 degrees f 0 degrees f
since water freezes at 0 degress celsius the temperature would have to be 0 degress or below.
Ice forms at 32 degrees Fahrenheit, which is 0 degrees Celsius.
Water is a solid at all temperatures from 0 degrees C downwards. That is at normal atmospheric pressure.
Yes, it is correct.
The density of water at 100 degrees Celsius is about 0.9584 grams per cubic centimeter. At this temperature, water is in its liquid state and expands slightly compared to when it is at its maximum density at 4 degrees Celsius.
The answer depends on the amount of water vapour in the air.
The amount of propane needed to heat a 50 gallon water heater from 75 degrees to 105 degrees would depend on factors such as the efficiency of the water heater, the starting temperature of the water, and the ambient temperature. It is difficult to provide an exact amount without this information.
A calorie is the amount of heat you need to raise the temperature of one gram of water by one degree Celsius. Assuming you are raising the temperature of the water from twenty degrees Celsius to ninety-nine degrees Celsius, it would take 20,000 calories. To calculate this, subtract 20 from 99. This is the amount of degrees you need to raise the temperature of the water by. Then multiply that number by 256, the amount of water in grams. You should get 20,244 calories. In significant digits, your answer should be 20,000 calories.
it freezes at 0C and boils at 100C
less than 32 degress ferhanhite