38.2C-24.5C = 13.7 C change
this requires 13.7 calories per gram or 1370 calories total
q (amt of heat) = mass * specific heat * temp. differenceThe specific heat of water is 1.00 cal/goC & the temperature difference is 70-30 = 40oCq = (105 grams)*(1.00 cal/goC)*(40oC)= 4,200 calories
Raise the temp of 52 grams of water from 33.0 C to 100 C = 52*67*4.184 = 14.577 kJConvert evaporate 52 g of water to steam without change of temp = 52*2259 = 117.468 kJRaise the temp of 52 grams of steam from 100 C to 110 C = 52*10*2.02 = 1.051 kJTotal energy required = 133.095 kJ = 31,811 calories or 31.811 kCal.
That completely depends on the specific heat capacity of the substance of which the sample is composed, which you haven't identified. It will take a lot more heat energy to raise the temperature of 65 grams of water 35 degrees than it would take to do the same to 65 grams of iron or yogurt, e.g.
1 gm/ml
100
1370 calories
To raise 1000 grams of water from 50 to 100 degrees requires 50 degrees x 1000 grams of heat, so the answer is 50,000 calories. Water at 100 degrees requires an additional 550 calories to convert 1 gram fully into steam. Therefore the remaining 50,000 calories can convert 50,000/550 grams into steam. So 90.9 grams become steam, and that's the answer.
700
105C
(5)(3)= 15 calories. 1 calorie is the energy (heat) to raise 1 gram of water by 1 degree celsius, so 5 grams of water (3 degrees Celsius) = 15.
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.
one calorie of heat is able to raise one gram of water one degree Celsius so 400 calories could raise 1g of water 400 degrees, so it would raise the 80g by(400/80) 5 degrees Celsius plus the initial temp of 10 degrees, the 80g of water would have a final temp of 15 degrees Celsius
46 calories (or 192, 464 joules) for each Celsius degree.
21 grams through 71 degrees is 21x71 calories.
1,000 grams of water by 75 degrees Celsius
46 calories (or 192, 464 joules) for each Celsius degree.
q (amt of heat) = mass * specific heat * temp. differenceThe specific heat of water is 1.00 cal/goC & the temperature difference is 70-30 = 40oCq = (105 grams)*(1.00 cal/goC)*(40oC)= 4,200 calories