The human body converts internal chemical energy into work and heat at rates of 60 to 125W (called the basal metabolic rate). This energy comes from food and is usually measured in kilocalories [1 kcal = 4.186 kJ]. (Note that the nutritional ‘Calorie’ listed on packaged food actually equals 1 kilocalorie). How many kilocalories of food energy does a person with a metoblic rate of 103.0 W require per day? Enter the numerical answer without units.

## The human body converts internal chemical energy into work and heat at rates of 60 to 125W (called the basal metabolic rate). This energy comes from food and is usually measured in kilocalories [1 kcal = 4.186 kJ]. (Note that the nutritional ‘Calorie’ listed on packaged food actually equals 1 kilocalorie). How many kilocalories of food energy does a person with a metoblic rate of 103.0 W require per day? Enter the numerical answer without units.

The human body converts internal chemical energy into work and … Read More...
Q explain our sun makes heat and light. Describe the process , including initial and final material.

## Q explain our sun makes heat and light. Describe the process , including initial and final material.

The change in the internal energy of a system that absorbs 2,500 J of heat and that has received 7,655 J of work by the surroundings is __________ J. A) -10,155 B) -5,155 C) 7 −1.91×10 D) 10,155 E) 5,155

## The change in the internal energy of a system that absorbs 2,500 J of heat and that has received 7,655 J of work by the surroundings is __________ J. A) -10,155 B) -5,155 C) 7 −1.91×10 D) 10,155 E) 5,155

D) 10,155
c. A wall with an area 0f 35 m2 is made up of six layers. On the inside is plaster 20 mm thick, then there is the brick 100 mm thick, followed by the insulation of 60 mm thick, then the brick of 100 mm thick, then there is the insulation of 65 mm thick and finally brick 100 mm thick. Calculate the thermal resistance, the heat transfer between the layers and the overall heat transfer coefficient given that the thermal conductivity of plaster is 20 W/ m K, the thermal conductivity of the brick is 0.6 W/m K and the thermal conductivity of the insulation is 0.08 W/ m K. The inner surface temperature of the wall is 22oC and the outer is -4oC.

## c. A wall with an area 0f 35 m2 is made up of six layers. On the inside is plaster 20 mm thick, then there is the brick 100 mm thick, followed by the insulation of 60 mm thick, then the brick of 100 mm thick, then there is the insulation of 65 mm thick and finally brick 100 mm thick. Calculate the thermal resistance, the heat transfer between the layers and the overall heat transfer coefficient given that the thermal conductivity of plaster is 20 W/ m K, the thermal conductivity of the brick is 0.6 W/m K and the thermal conductivity of the insulation is 0.08 W/ m K. The inner surface temperature of the wall is 22oC and the outer is -4oC.

1) R= R1+ R2+ R3+ R4+ R5+ R6             = … Read More...
Determine the changes in enthalpy, internal energy and entropy when 2.7 kg of water taken at P1 = 1.0133 X 10^5 Pa and T1 = 293 K evaporate at P2 = 0.50665 X 105 Pa and T2 = 373 K . Cp(l) ~ Cv(l) = 4.187 X 10^3 J/kg/K the specific heat of evaporation being 2260.98 X 10^3 J/kg =Spv

## Determine the changes in enthalpy, internal energy and entropy when 2.7 kg of water taken at P1 = 1.0133 X 10^5 Pa and T1 = 293 K evaporate at P2 = 0.50665 X 105 Pa and T2 = 373 K . Cp(l) ~ Cv(l) = 4.187 X 10^3 J/kg/K the specific heat of evaporation being 2260.98 X 10^3 J/kg =Spv

Soultion 1)    delta H = m*Cv*DeltaT + m*Spv   = 2.7*4.187*(373-293)+2.7*2260.98 … Read More...

Determine the changes in enthalpy , internal energy and entropy when 2.7 kg of water taken at P1 = 1.0133 X 10^5 Pa and T1 = 293 K evaporate at P2 = 0.50665 X 10^5 Pa and T2 = 373 K . Cp(l) ~ Cv(l) = 4.187 X 10^3 J/kg/K the specific heat of evaporation being 2260.98 X 10^3 J/kg

## Determine the changes in enthalpy , internal energy and entropy when 2.7 kg of water taken at P1 = 1.0133 X 10^5 Pa and T1 = 293 K evaporate at P2 = 0.50665 X 10^5 Pa and T2 = 373 K . Cp(l) ~ Cv(l) = 4.187 X 10^3 J/kg/K the specific heat of evaporation being 2260.98 X 10^3 J/kg

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b. Explain with appropriate examples the difference between heat conduction, heat convection and heat radiation

## b. Explain with appropriate examples the difference between heat conduction, heat convection and heat radiation

CONDUCTION-– The transfer of heat among substances that are in … Read More...
Cv, the constant volume specific heat is defined as: