q=(275)(11)(1.00)
q=3,025 cal
remember that the formula to find heat is:
q=m(DT)Cp
and remember that the specific heat of water is: 1.00 cal/(gxC)
you just replace values
hope this helps:)
334.8 Joules
No heat (energy) is required to freeze water (from liquid to solid). Freezing RELEASES energy (heat), as it is an exothermic event. If you want to know how much energy is release, you need to know the heat of fusion for water, and then multiply that by the mass of water being frozen.
when we sweat , water use the body heat for vaporisation as water to vapour formation is a endothermic reaction. so the heat required for this reaction in case of sweat evaporation is body heat. as our body lose heat, our body becomes cool.
The specific heat of a substance allows us to calculate the amount of heat energy required to change its temperature. Water has a specific heat nearly 11 times great than copper, therefore, water will take 11 times more energy to heat. Also water heats slowly and copper heats and cools rapidly.
what is the temperature at which a substance changes from a solid to a liquid called
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
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.
No heat (energy) is required to freeze water (from liquid to solid). Freezing RELEASES energy (heat), as it is an exothermic event. If you want to know how much energy is release, you need to know the heat of fusion for water, and then multiply that by the mass of water being frozen.
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.
The necessary heat is 1,45 kcal.
Heat is not required for water condensation. Water vapors condenses into water droplets.