First, you need to melt the ice. Look up the heat of fusion of ice, and multiply that by the amount of grams.Then you need to heat the water, from zero degrees to 78 degrees. Look up the specific heat of water, and then multiply together all of the following: The specific heat; the mass; the temperature difference.
Finally, add the two together.
The heat energy required to melt ice at 0 degrees Celsius is called the heat of fusion. For ice, the heat of fusion is approximately 334 J/g. To convert this to calories, divide by 4.184 J/cal, which gives you approximately 80 calories of heat energy needed to melt 10 grams of ice.
This heat is 51, 33 cal.
To melt ice, the energy required is known as the heat of fusion, which is 334 J/g. To convert this to calories, we need to divide by 4.184 J/cal. So, the total calories required to melt a 25g ice cube at 0 degrees Celsius would be 25g x 334 J/g ÷ 4.184 J/cal = 1991 calories.
The amount of calories needed to heat up a substance can be calculated using the specific heat capacity of the substance. For water, the specific heat capacity is 1 calorie/gram°C. A 2-liter bottle of water weighs around 2000 grams. To raise the temperature of 2000 grams of water from 0°C to 100°C, you would need approximately 200,000 calories.
Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram. So, the total number of calories in the food would be (20 grams carbohydrate x 4) + (8 grams protein x 4) + (5 grams fat x 9) = 80 + 32 + 45 = 157 calories.
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
To melt 10 grams of ice at 0 degrees Celsius, it would require 80 calories of heat energy per gram, so a total of 800 calories (80 calories/gram * 10 grams = 800 calories) would be needed.
(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.
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
1370 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.
The heat energy required to melt ice at 0 degrees Celsius is called the heat of fusion. For ice, the heat of fusion is approximately 334 J/g. To convert this to calories, divide by 4.184 J/cal, which gives you approximately 80 calories of heat energy needed to melt 10 grams of ice.
100
The amount of heat needed to raise the temperature of a substance is given by the formula: Q = mcΔT, where Q is the heat (in calories), m is the mass (in grams), c is the specific heat capacity of water (1 cal/g°C), and ΔT is the change in temperature. Plugging in the values: Q = 8g * 1 cal/g°C * 7°C = 56 calories.
You use Heat of fusion... Heat=mass x heat of fusion Heat of fusion for water: 80 cal/g so 35g x 80 cal/g= 2800 cal released.
How many calories are in 26 grams of carbohydrates?"