q (amt of heat) = mass * specific heat * temp. difference
The specific heat of water is 1.00 cal/goC & the temperature difference is 70-30 = 40oC
q = (105 grams)*(1.00 cal/goC)*(40oC)
= 4,200 calories
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
For one gram of ice, it takes 11.9 calories to change the temperature to 0°, 80 calories to melt the ice, 100 calories to raise the water temperature to 100°, 540 calories to change the water to steam, and 23 calories to raise the steam temperature to 123°. That's a total of (11.9 + 80 + 100 + 540 + 23) calories or 754.9 calories. So to do the same to 55.6 grams of ice requires 55.6 times as much heat. 754.9 calories times 55.6 equals approximately 41972 calories (about 42 kilocalories).
Density = grams/ml 1.00 g/ml = X g/5.0 ml = 5.0 grams water ============== q(joules) = mass * specific heat * change in temp. q = (5.0 grams)(4.180 J/gC)(75 C - 2.50 C) = 1515.25 Joules ---------------------------------/4.184 = 362 calories -------------------
38.2C-24.5C = 13.7 C change this requires 13.7 calories per gram or 1370 calories total
If the ice starts at 0 degrees Celsius and vaporization takes place at 100 degrees Celsius, then ... 10 lbs = 4540 grams Q = 80(4540) + 1(100 - 0)(4540) + 540(4540) Q = 720(4540) = 3,268,800 calories
700
105C
q=mass * Cs * delta T = 0.25kg * 4.18 * (85-10) =78J
21 Kg = 2100 grams to rise the temperature of this amount of water by 2 degrees Celsius you need 2*2100 = 4200 calories or 17572.8 Joules.
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
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.
80.5 calories 35-12=23 23*3.5=80.5 1c raises 1 gr. h2o 1degree centigrade Here is the formula, it should help a lot:Total Number of Calories = (Specific Heat of Water) ×(Mass of Water) × (Absolute Temperature Change)
1 calorie is needed to raise 1 g of water 1 °C. 350 * 22 = 7700 calories ■
1,000 grams of water by 75 degrees Celsius
103 calories. The heat of fusion of water is 80 cal/g and it takes one calorie to change the temperature of 1 g of water by 1 degree. 80+23=103 calories
For one gram of ice, it takes 11.9 calories to change the temperature to 0°, 80 calories to melt the ice, 100 calories to raise the water temperature to 100°, 540 calories to change the water to steam, and 23 calories to raise the steam temperature to 123°. That's a total of (11.9 + 80 + 100 + 540 + 23) calories or 754.9 calories. So to do the same to 55.6 grams of ice requires 55.6 times as much heat. 754.9 calories times 55.6 equals approximately 41972 calories (about 42 kilocalories).
21 grams through 71 degrees is 21x71 calories.