during an extended wilderness hike you develop a terrific craving for a frozen d
ID: 2217046 • Letter: D
Question
during an extended wilderness hike you develop a terrific craving for a frozen daiquiri, but have no ice. unfortunately, the temperature only drops to 6.0 c each night. however, you decide to try to make ice by exposing a shallow insulated dish of 4.5 g of water with a surface area = 9.0 cm2, depth = 5.0 mm, e = 0.90, and initial temperature = 6.0 c to the clear moonless sky which acts like a blackbody radiator at Ts = -23 c. find the time for the water to freeze by radiation. can the freezing be accomplished in one night? (note: this method was actually used before refrigerators were common)Explanation / Answer
Considering moon as a black body, and from Stefan–Boltzmann law, j(heat radiated per unit areaper unit time)=sT4, T is in Kelvin, s=5.67×10-8 W m-2 K-4 Therefore, Heat absorbed by the container will be esT4tA, where t is the time of exposure and A is the area exposed. (where e is the emmisivity which is equal to absorbity by Kirchoff's law of thermal radiation) Now we calculate the amount of heat given out for conversion of liquid water to solid ice, H=mc?T+mL, m is the mass of substance(here it is water), c is the specific heat capacity of the subtance (for water which is 4.1855 J/(g·K)) and ?t is the change in temperature and L is latent heat of fusion(334J/g) Equating both, mc?T+mL=esT4tA [4.5*4.1855*(6)]+[4.5*334]=0.9*5.67×10-8 *{(273-23)^4}*9*10^(-6)*t Solving we get, t=250.19 hours Considering moon is visible for 10 hours in a day, it will take approximately 25 days.
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