Vitamins do not have any calories.
"calories" (lowercase C) and "kilocalories" or "kcal" are interchangeable words. 1 Calorie (uppercase C) = 1000 calories (lowercase C) = 1 kilocalorie So, 390 kcal is equal to 390 calories as well.
The specific heat capacity of aluminum is 0.897 J/g°C. To convert this to kcal/g°C, we divide by 4.184 to get 0.214 kcal/g°C. Therefore, the total kilocalories of heat required to raise the temperature of 225g of aluminum from 20°C would be 225g * 20°C * 0.214 kcal/g°C = 966 kcal.
To calculate the total kilocalories needed, first, you need to melt the ice. The latent heat of fusion for ice is approximately 80 kcal/kg, so melting 0.12 kg of ice requires 0.12 kg × 80 kcal/kg = 9.6 kcal. After melting, you need to raise the temperature of the resulting water from 0°C to 25°C. The specific heat of water is about 1 kcal/kg°C, so heating 0.12 kg of water by 25°C requires 0.12 kg × 25 kcal/kg°C = 3 kcal. Therefore, the total heat needed is 9.6 kcal + 3 kcal = 12.6 kcal.
Hi The answer to your question is there is 3500 kcal in one pound!
1000 calories in one kcal so 1000 is to 1 as x is to 60 = .06 kcal
760 kcal/ton clinker
There are approximately 7.7 kcal in 1 gram of fat, 4 kcal in 1 gram of protein, and 4 kcal in 1 gram of carbohydrates. The exact number of kilocalories (kcal) in 1 gram depends on the specific macronutrient being measured. For alcohol, there are about 7 kcal per gram.
To convert W/mK (watts per meter Kelvin) to kcal/mh°C (kilocalories per meter hour Celsius), you can use the following formula: 1 W/mK = 8.597 kcal/mh°C. Using the conversion factor, you can multiply the value in W/mK by 8.597 to obtain the equivalent value in kcal/mh°C.
1 kCal = 1000 Cal or I Cal = 1/1000 kCal
since 1 pound = 3500 kcal ... then 145 pounds = 507500 kcal (145 x 3500) hope i helped :D
Fat free feta has 40 kcal for 1/4 c.
if i need 4500x103 kcal/heat to heat something how many litres of LPG per hr is that