## Answered) 1) A 0.250-kg aluminum bowl holding 0.800 kg of soup at 25.8 C is placed in a freezer. What is the final temperature if 428 kJ of energy is

1) A 0.250-kg aluminum bowl holdingÂ 0.800 kgÂ of soup atÂ 25.8Ã‚Â°C is placed in a freezer. What is the final temperature ifÂ 428Â kJÂ of energy is transferred from the bowl and soup? Assume the soup has the same thermal properties as that of water, the specific heat of the liquid soup isÂ 1.00 kcal/(kgÂ Ã‚Â·Â Ã‚Â°C),Â frozen soup isÂ 0.500 kcal/(kgÂ Ã‚Â·Â Ã‚Â°C),Â and the latent heat of fusion isÂ 79.8 kcal/kg.Â The specific heat of aluminum isÂ 0.215 kcal/(kgÂ Ã‚Â·Â Ã‚Â°C).

2) A one-story house is heated with natural gas, which gives offÂ 3.890Â Â 107Â JÂ of energy for every cubic meter of gas burned. Take the average temperature difference between the inside and outside for the 120-day heating season to beÂ 18Ã‚Â°C.Â In the diagram belowÂ h1Â = 5.00 m,Â WÂ = 8.00 m,Â LÂ = 10.0 mÂ andÂ h2Â = 3.00 m.

(a) If the walls, windows, and roof have an average thickness ofÂ 22.0Â cmÂ and the thermal conductivity (including the windows) isÂ 0.48 W/(mÂ Ã‚Â·Â Ã‚Â°C),Â determine how many cubic meters of gas is used up during the 120-day winter season? Assume air infiltration and heat loss through the ground is negligible.

(b) Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as two years.) Suppose you wish to install an extraÂ 8 cmÂ of fiberglass to the roof. If energy costs \$0.90 per one hundred million joules and the insulation was \$4.00 per square meter, then calculate the simple payback time taking the 120-day heating season to be equivalent to one year.

3) The Sun radiates like a perfect blackbody with an emissivity of exactly 1.

(a) Â Calculate the surface temperature of the Sun, given it is a sphere with aÂ 7.00Â Â 108Â m radius that radiatesÂ 3.80Â Â 1026Â WÂ intoÂ 3 KÂ space.

(b) How much power does the Sun radiate per square meter of its surface?

(c) Â How much power in watts per square meter is this at the distance of the Earth,Â 1.50Â Â 1011Â m away? (This number is called the solar constant.)

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