You are in charge of keeping drinks cold for a picnic. You have a styrofoam (10^

Question

You are in charge of keeping drinks cold for a picnic. You have a styrofoam (10^-2 w/mK) box that is filled with cans of cola and you plan to put some 0 degree C ice in it. It will sit in a shady corner of the park all day. Your task is to buy enough ice to put in the box at 6am so that the temperature stays at 0 degrees C until the picnic starts at 6pm. you don't want to buy much ice because that means that you'll have less money to spend on food and other picnic items. You plan to keep six cold two liter bottles of cola (which has the same heat capacity as water). Over night you will keep the drinks in the fridge so they will start out at about 0 degress C.

How much ice will you need?

Today will be a beautiful sunny day, a perfect day for a picnic. The temperature (i.e. in the shade) will start out at 10 degrees C and increase at a steady rate reaching 35 degrees C at 6pm. The cooler is roughly cubical with edge length 50 cm and wall thickness 2cm. You can assume that the ground on which the box rests is at a constant temperature of 15 degrees C.

You start by making the simplifying assumption that everything inside the cooler is ast 0 degrees C. Just before you buy the ice, you realize that assuming the entire inside of the cooler is at a constant temperature of 0 degrees C is equivalent to assuming that the inside of the cooler is always in a state of thermal equilibrium. This makes you worry what would happen if this assumption failed to be true. How could a deviation from thermal equilibrium inside the cooler affect the amount of ice required?

(Data: heat of fusion for water= 344 kJ/kg, heat of vaporization for water=2256 kJ/kg, specific heat for water=4186J/(kg K), specific heat for ice= 2100J/(kg K), emissivity of water=0.98, emissivity of ice=0.98)

I need all equations used, any laws used, and any algebra used. Thanks.

Explanation / Answer

Here, total heat energy absorbed = heat of fusion + heat to raise temperatue of ice + heat to raise temperatue of water   + heat to vapourize water

=>   total heat energy absorbed =   m * 344000 + m * 2100 * 15 + m * 4186 * 35 + m * 2256000

=>                      1.56 * 108         =   m * 2778010

=>   m   =     56.155 kg     of ice will be needed .

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