Want this question answered?
how much heat is required to convert 0,3kg of ice at 0c to water at the same tempture
Heat required = mass x specific heat of water x temperature difference Here we have heat required = 21 x 1 x 10 = 210 cals
heat
You need to add all of the following:* The heat required to heat ice from -5 to 0 degrees. Multiply the mass times the temperature difference times the specific heat of ice. * The heat required to melt ice. Multiply the mass by the heat of fusion. * The heat required to raiste the temperature of water from 0 to 20 degrees. Multiply the mass times the temperature difference times the specific heat of water.
The molar heat of fusion of water in J / g is 334. To find the heat required to convert 0.3 kg, use the equation: heat of fusion * mass = heat required. It would require 100.2 kJ.
Its a non polar compound that will take less energy to break the bonds
it takes 2 pounds of it
80 calories of heat are required to melt one gram of ice without altering the temperature. There are 2267.96 grams in five pounds so you would need 181436.8 calories of heat to melt all that ice without raising the temperature.
how much heat is required to convert 0,3kg of ice at 0c to water at the same tempture
Heat
Heat required = mass x specific heat of water x temperature difference Here we have heat required = 21 x 1 x 10 = 210 cals
Heat required for this transition is given as the the sum of three heatsheat required for heating the ice from -5 degree Celsius +latent heat(conversion of ice at zero degree to water at zero degrees)+heat required to heat the water from 0 to 5 degree CelsiusHeating of ice=m x s x delta T,where m is the mass ,s is the specific heat of ice=200x0.5x5=500calmelting of ice=mxlatent heat=200x80=16,000calHeating of water=m x s x delta T,where m is the mass ,s is the specific heat of water =200x1x5=1000calTotal heat required=500+16,000+1000=17,500 cal
heat
You need to add all of the following:* The heat required to heat ice from -5 to 0 degrees. Multiply the mass times the temperature difference times the specific heat of ice. * The heat required to melt ice. Multiply the mass by the heat of fusion. * The heat required to raiste the temperature of water from 0 to 20 degrees. Multiply the mass times the temperature difference times the specific heat of water.
Any external heat source can provide the required energy.
334.8 Joules
Large amount of heat is required by water to increase the temperature of water. In addition, the heat of vaporization is required to tern the water into the vapour. This property of water helps the animal to keep it cool by use of less amount of water.