The temperature would be that of water's boilng point od 100 degrees
I believe it will be 145.52 degrees Celsius if I did my math correctly. You need to convert calories to joules. I believe one joule raises the temp of 1 gram water by 1 degree Celsius so 1200*4.184=5020.8 J /40grams=125.52 temp increase+20=145.52 degrees Celsius.
The density of sulfur in grams/cm3 is 2.070. (not at twenty five degrees Celsius)
At room temperature, mercury (Hg) is a liquid, and methane (CH4) is a gas. The melting point of methane is -182.5 degrees Celsius. The melting point of mercury is -39 degrees Celsius. It takes more heat to melt the mercury.
what is the molecular mass of 1-propanol
32 g KCl
If coffee and milk have the same thermal coefficient then: (15*22 + 185*86)/200 = 81 (81,2 rounded off due to significant digits)
Approx 4974 Joules.
I'll assume here that by "70 temperature" you mean "70 degrees Celsius". Basically, you have to calculate the average temperature of all of the water in the mixture, which will be the final temperature once it's well stirred. The 200 grams of water at 10 degrees represent 2/3 of the total amount of water (300 grams), so thus, multiply 10 by 2/3 to determine their contribution to the final temperature. You will get 20/3. The 100 grams of water at 70 degrees represent 1/3 of the total amount of water, so multiply 70 by 1/3 to determine their contribution to the final temperature. You will get 70/3. When you add together the two temperatures you get 90/3, which is equal to 30. Therefore, the final temperature is 30 degrees Celsius.
Liters measure volume. Grams are a measure of mass, degrees Celsius are a measure of temperature, and meters are a measure of length.
105C
Density of ice at 0 degrees Celsius is 916.8 grams per cubic centimeter or milliliter. The density of fresh water is dependant on the temperature: At 3.98 degrees Celsius the density is 0.999975 grams per milliliter. At 100 degrees Celsius the density is 0.958.35 grams per milliliter.
Do some converting first. 688 calories (4.184 Joules/1 calorie) = 2878.592 Joules 25 ml of water = 25 grams q(Jolules) = mass * specific heat * (Temp. final - Temp. initial) 2878.592 Joules = 25 grams Water * 4.180 J/gC * (Temp Final - 80C ) 2878.592 Joiles = 104.5( Temp. Final) - 8360 11238.592 =104.5(Temp. Final) 107.55 Celsius Final Temperature ( call it 108 C )
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.
No, grams are unit of mass, not temperature. Temperature is measured in °C (degrees celsius) or for scientific work in 'K' K = °C + 273.15
q(joules) = mass * specific heat * change in temperature q = (500 grams H2O)(4.180 J/goC)(100o C - 20o C) = 1.7 X 105 joules ================add this much heat energy to the water
1,000 grams of water by 75 degrees Celsius
I believe it will be 145.52 degrees Celsius if I did my math correctly. You need to convert calories to joules. I believe one joule raises the temp of 1 gram water by 1 degree Celsius so 1200*4.184=5020.8 J /40grams=125.52 temp increase+20=145.52 degrees Celsius.