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exam6 solutions - Problem 1[20 points Your room on the...

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Problem 1. [20 points] Your room on the lower deck of cruise liner is below sea level and is heated to a comfortable 25°C. The water outside is 5°C. One of your walls is the hull of the ship. The hull is 5cm steel with a thermal coefficient of 15 W/m-°C. The wall area is 12m 2 . You can ignore the ceiling, floor and other three walls of your room in this problem. (a) What is the rate of heat transfer through this wall? Solution: Since there is no sink or source through this wall, the rate of heat transfer is a constant. dQ dt =− kA dT dx = Const → ( dQ dt ) 0 d wall dx =− kA T room T water dT Therefore, dQ dt =− kA T water T room d wall = 72 kW (b) Your heater is a heat pump operating at its ideal Carnot performance. How much power must be supplied to the heater? Solution: The assumed ideal Carnot performance indicates that η = ́ Q room ́ W = T room T room T water ,i.e., ́ W = d Q room dt ( 1 T water T room ) Where ́ Q room stands for the heat released per unit time to the room, which possesses a higher temperature. In order to keep the temperature of the room constant, we require d Q room dt = dQ dt Therefore, ́ W = dQ dt ( 1 T water T room ) = 72kW × ( 1 278 298 ) = 4.83 kW
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Problem 2 . [20 points] A hot rock of mass m=10 kg at a temperature T=1500 K is placed into a container which is filled with 1m 3 of water at a temperature of 300 K. The specific heat of water can be approximated as 4kJ/kg-K and the specific heat of the rock is 1kJ/kg-K. The density of water is 1g/cm 3 . (a) What is the final temperature of the rock and water? (b) What is the entropy change of the rock? (c) What is the entropy change of the water? (d) By how much did the height of the water change due to heating alone (ignoring the bigger effect of the addition of the rock) if the container has a depth of 2m and a surface area of 0.5m 2 ? The coefficient of volume expansion of water is 2x10 -6 /°C. Solutions: (a) Assume that the final temperature is T f . c R m ( T f T ) + c w m w ( T f T w ) = 0 T f = c R mT + c w m w T w c R m + c w m w = 303 K (b) Δ S rock = T T f c R md T ' T ' = c R mln ( T f T ) = 10ln ( 303 1500 ) kJ / K =− 16 kT / K (c) Δ S water = T w T f c w m w dT T = c w m w ln ( T f T w ) = 4 × 10 3 ln ( 303 300 ) kJ / K = 40 kT / K (d) The change of the volume of the water according to the thermal expansion is ΔV = β V ( T f T w ) = 6 × 10 6 m 3 Therefore, the change of the height is Δh = ΔV Α = 6 × 10 6 m 3 0.5 m 2 = 1.2 × 10 5 m
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Problem 3. [14 points] A thunderstorm gets energy by condensation of water vapor in the air. Suppose that the storm condenses all the water vapor in 10km 3 (ten cubic kilometers) of air. Assume that at the conditions of the storm there is 100% relative humidity and that this corresponds to .02 kg of water in every cubic meter and a heat of vaporization of 2x10 3 kJ/kg.
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