If you assume that the density of sweat is the same as that of water (1g/ml) and that the specific heat of sweat is the same as that of water (4.184 J/g/deg), then one can estimate the heat required to evaporate 7.24 ml of sweat at 25 degrees Cheat to raise 7.24 ml from 25º to 100º = q = mC∆T = (7.24g)(4.184 J/g/deg)(75deg) = 2272 J
heat to evaporate 7.24 ml = q = m∆Hvap = (7.24g)(2260 J/g) = 16,362 J
Total heat = 2272 J + 16,362 J = 18,634 J = 18.6 kJ
To calculate the heat needed to vaporize sweat, you would need to know the specific heat of vaporization of sweat. Once you have that information, you can use the formula Q = mL, where Q is the heat needed, m is the mass of the substance (converted from volume using its density), and L is the specific heat of vaporization.
The first step is to determine the mass of 7.24 ml of sweat by using its density of 0.9970 g/ml. The mass can be calculated as 7.24 ml * 0.9970 g/ml = 7.22 g. Next, we can use the heat of vaporization of water, which is 2260 J/g, to find the total heat required. Therefore, the amount of heat needed to vaporize 7.24 ml of sweat would be 7.22 g * 2260 J/g = 16367.2 J.
The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.
1650kj
The heat required to vaporize 500 grams of ice at its freezing point is the sum of the heat required to raise the temperature of the ice to its melting point, the heat of fusion to melt the ice, the heat required to raise the temperature of water to its boiling point, and finally the heat of vaporization to vaporize the water. The specific heat capacity of ice, heat of fusion of ice, specific heat capacity of water, and heat of vaporization of water are all needed to perform the calculations.
To calculate the heat needed to vaporize sweat, you would need to know the specific heat of vaporization of sweat. Once you have that information, you can use the formula Q = mL, where Q is the heat needed, m is the mass of the substance (converted from volume using its density), and L is the specific heat of vaporization.
cotton allows sweat to vaporize and it absorbs heat.
The first step is to determine the mass of 7.24 ml of sweat by using its density of 0.9970 g/ml. The mass can be calculated as 7.24 ml * 0.9970 g/ml = 7.22 g. Next, we can use the heat of vaporization of water, which is 2260 J/g, to find the total heat required. Therefore, the amount of heat needed to vaporize 7.24 ml of sweat would be 7.22 g * 2260 J/g = 16367.2 J.
The amount of time depends on the latent heat of vaporisation and the latent heat of melting.
The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.
Gasoline has a less boiling point (72 degree Celsius) While water has a bigger boiling point (100 degree Celsius)
1650kj
1oo calories for 1 g
The heat required to vaporize 500 grams of ice at its freezing point is the sum of the heat required to raise the temperature of the ice to its melting point, the heat of fusion to melt the ice, the heat required to raise the temperature of water to its boiling point, and finally the heat of vaporization to vaporize the water. The specific heat capacity of ice, heat of fusion of ice, specific heat capacity of water, and heat of vaporization of water are all needed to perform the calculations.
The amount of energy needed to vaporize one gram of a liquid is known as the heat of vaporization. It varies for different substances and can be calculated using the formula Q = m * Hv, where Q is the energy required, m is the mass of the substance (in this case, 1 gram), and Hv is the heat of vaporization of the specific substance.
you sweat to stay cool. The heat of the sun is used to turn your sweat to water vapor, rather than to heat up your body.
The heat of vaporization of ethanol is approximately 840 kJ/kg. To find the total heat required to vaporize 1.25 kg of ethanol, you can multiply the mass by the heat of vaporization: 1.25 kg * 840 kJ/kg = 1050 kJ.