here , you multiply the the mass of the ice times the latent heat of fusion so the answer is
5*3.33*(10^5)*(10^-3)=1665 Joule
so to convert the result into calories divide by 4.186(1 calorie -------------->4.186 Joule)
the result is : 397.75 Calorie
700
1 calorie is needed to raise 1 g of water 1 °C. 350 * 22 = 7700 calories ■
The answer is 2 calories.
100 calories. 1 calorie is defined as the amount of energy required to raise the temperature of 1 gram of 1 by 1 degree Celsius. So, if you need to raise 10 grams of water 1 degree, you would need 10 calories of energy. If you needed to raise those same 10 gram by 10 degrees, you'll need 10 * 10, or 100 calories.
1 calorie is defined as the amount of energy needed to raise the temperature of 1 gram of water by 1C, so... It takes 8.1 calories to raise your 8.1 grams by 1C, but you need to raise it 20C. 8.1*20=162. 162 calories is the answer you are looking for.
69
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
calories were never "made." they are simply the amount of energy required to raise the temperature one gram of water one degree celsius.
1 calorie is the energy required to raise 1 gram of water by 1 degree C. So it would take 5 calories to raise it by 5 degrees C.
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).
The number of calories required will depend on the mass of water which is to be heated.
specific heat(; your welcome!