In exercising, a weight lifter loses 0.124 kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is 1.44E+5 J. Assuming that the latent heat of vaporization of perspiration is 2.41E+6 J/kg, find the change in the internal energy of the weight lifter.
In exercising, a weight lifter loses 0.124 kg of water through evaporation, the heat required to...
In exercising, a weight lifter loses 0.174 kg of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is 1.40E+5 J. Assuming that the latent heat of vaporization of perspiration is 2.40E+6 J/kg, find the change in the internal energy of the weight lifter.
When you jog, most of the food energy you burn above your basal metabolic rate (BMR) ends up as internal energy that would raise your body temperature if it were not eliminated. The evaporation of perspiration is the primary mechanism for eliminating this energy. Determine the amount of water you lose to evaporation when running for 46 minutes at a rate that uses 400 kcal/h above your BMR. (That amount is often considered to be the "maximum fat-burning" energy output....
How much thermal energy (in J) is required to boil 2.20 kg of water at 100.0°C into steam at 143.0°C? The latent heat of vaporization of water is 2.26 ✕ 106 J/kg and the specific heat of steam is 2010 J / kg · °C . J
Calculate by the energy balance method the evaporation rate in [mm day] from an open water surface, if the net radiation is 225[Wrm] and the air temperature 22°C, assuming no sensible heat or ground heat flux, the water density is 997 kg/m, and the latent heat Jkgl2.501x10-2370T where Temperature T is in [C]. Is your estimation reasonable in reality? [15pt]
Calculate by the energy balance method the evaporation rate in [mm day] from an open water surface, if the net...
5) How much heat is required to warm 1.50 L of water from 25.0 °C to 100 °C? (Density of water = 1,0 g/mL for the water; Cs of water 4.18 J/g oC). Evaporating sweat cools the body because evaporation is an endothermic process which can be expressed using the following reaction: H20 (g) AHOxn +44.01 kJ H2O (I) Estimate the mass of water that must evaporate from the skin to cool the body by 0.5 °C. Assume a body...
An aluminum tea kettle of 0.45-kg containing 2.40-kg of water at
25°C is placed on a
burner. Assuming that the burner is rated at 2000 Watt, how long
will it take to bring
the water to a boil? (b) Once boiling begins, how much time is
required to boil all the
water out of the tea kettle? Assume that no energy is lost in this
process.
Latent Heat of Fusion Latent Heat of Vaporization Specific Heats SubstcJkg. C cal/g- CS...
21. How much heat is required to change m=1 kg of ice at -6oC into water at 60oC ? cw=4190 J/KgK(specific heat of water), Lf=333kJ/Kg (latent heat of fusion for ice), cice=200J/kgK (specific heat of ice).
Pleases help me thanks.
5 kg of liquid water at 100°C was converted to steam at 100°C by boiling at standard atmospheric pressure, in the arrangement of the figure below. The volume of the water changed from an initial value of 0.005 mº as a liquid to 8.355 mº as steam. Latent heat of vaporization for water Lv = 2256 kJ/kg. Standard atmospheric pressure is 1 atm or 1.01 x 109 Pa. Loaded Piston Steam Insulation Insulation Liquid water Thermal...
At 1 atm, how much energy is required to heat 0.0550 kg of ice at -22.0 °C to steam at 129.0 °C? STRATEGY 1. Calculate the energy needed for each temperature change or phase change individually. A. The energy needed to heat 0.0550 kg of ice from -22.0 C to its melting point. B. The energy needed to melt 0.0550 kg of ice at its melting point C. The energy needed to heat 0.0550 kg of liquid water from the...
The specific heat of water is 4186 J/kg C. How much does the internal energy of 300 grams of water change as it is heated from 23 to 46 degrees C? Assume that the volume is constant. Internal energy change = Number Joules