When allowed to stand for long enough, the final temperature will reach room temperature.
When allowed to stand for long enough, the final temperature will reach room temperature.
Essentially, because the room is a much bigger object than the water. This is a general consequence of entropy. The heat distribution in the entire system of the room and the water tends to 'average out' over time when left on its own. Since the room (or atmosphere) is much bigger the final temperature will be very close to its original temperature. This is also why water at zero degrees will heat up to room temperature
It depends on what your INITIAL and FINAL temperatures are. Assuming you had water at 20°C and wanted to raise it to 50°C You can use the formula: Q = mcΔT Q = Amount of Heat (Energy) m (mass)= 1g c (specific heat capacity) = 4.186J/g°C ΔT = Final Temperature - Initial Temperature = 50°C - 20°C = 30°C Q = 1g x 4.186J/g°C x 30°C Q = 125.58J
You can weigh it; or you can calculate it based on the known density of water.
I doubt that is possible to compress 98.5 g of water as to occupy only 50 ml. 50 ml of water would weigh about 50 g.
When allowed to stand for long enough, the final temperature will reach room temperature.
No, as both the temperatures are the same, you will get only 2 cups, each 50 degrees. You have to heat the cup to get 100 degree.
The temperature can be 50 to 60 degrees. water temperature
50 degrees.
Neither is hotter they both have a temperature of 50!
Essentially, because the room is a much bigger object than the water. This is a general consequence of entropy. The heat distribution in the entire system of the room and the water tends to 'average out' over time when left on its own. Since the room (or atmosphere) is much bigger the final temperature will be very close to its original temperature. This is also why water at zero degrees will heat up to room temperature
The cup of water is as hot as the bath tub full of water as they are at the same temperature.
It depends on what your INITIAL and FINAL temperatures are. Assuming you had water at 20°C and wanted to raise it to 50°C You can use the formula: Q = mcΔT Q = Amount of Heat (Energy) m (mass)= 1g c (specific heat capacity) = 4.186J/g°C ΔT = Final Temperature - Initial Temperature = 50°C - 20°C = 30°C Q = 1g x 4.186J/g°C x 30°C Q = 125.58J
U.S.gallon = 8.33 pounds of water. Therefore to raise the temperature by one degree F will require 8.33 BTU. The initial temperature of 50 F is inconsequential.
You can weigh it; or you can calculate it based on the known density of water.
i think 50% but the hotter the temperature is the more salt will dissolve
50 degrees or less year round