-28.12 deg F
334 ml = .334 litres
.334
167 multiplied by 2 is 334.
234+100=334
334mL = 0.334 liters.
To melt 1 gram of ice at 0°C, it requires 334 joules of energy. So for g grams of ice, the energy needed would be g multiplied by 334 joules.
To find out how many 11's are in 334, divide 334 by 11. When you perform the division, 334 ÷ 11 equals approximately 30.36. Since we're looking for whole instances of 11, there are 30 complete 11's in 334.
33,400 grams in one kg
334 metres is very much bigger.
The heat lost by water at 0 degrees Celsius to change to ice is equal to the heat of fusion of water, which is about 334 joules per gram. So, for 2 grams of water, the heat loss would be 2 * 334 = 668 joules.
To melt 25g of ice at 0 degrees Celsius, you need 334 J/g of energy (latent heat of fusion of ice). To convert this to kilojoules: 334 J/g * 25g = 8350 J = 8.35 kJ.
To freeze liquid water at 0 degrees Celsius, you need to remove 334 J/g of energy. To cool it from 0 degrees Celsius to -20 degrees Celsius, you need to remove an additional 80 J/g. Therefore, for 75g of water, the total energy released would be 75g * (334 J/g + 80 J/g) = 33,200 J.