And please dont just tell me the answer, I need help understanding how to figure this out.
If coffee and milk have the same thermal coefficient then: (15*22 + 185*86)/200 = 81 (81,2 rounded off due to significant digits)
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
The change in temperature is 25 degrees Celsius, meaning it takes 22.48 joules per degree of change. The specific heat of iron is 0.449 J/g degree Celsius. This means that the mass of iron must be 50.07 grams
The answer depends on the quantity of air.
700
The temperature would be that of water's boilng point od 100 degrees
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.
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.
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.
Fresh water at a temperature of 10 degrees Celsius has a density of 999.70 kilograms per cubic meter or 0.9997 grams per cubic centimeter. Sea water with a salinity of 3.5 percent at a temperature of 10 degrees Celsius has a density of 1026.98 kilograms per cubic meter or 1.02698 grams per cubic centimeter.