To help prevent frost damage, fruit growers sometimes protect their crop by spra

Question

To help prevent frost damage, fruit growers sometimes protect their crop by spraying it with water when overnight temperatures are expected to go below the freezing mark. When the water turns to ice during the night, heat is released into the plants, thereby giving them a measure of protection against the falling temperature. Suppose a grower sprays 7.19 kg of water at 0^degree C onto a fruit tree. How much heat is released by the water when it freezes? How much would the temperature of a 182- kg tree rise if it absorbed the heat released in part(a)? Assume that the specific heat capacity of the tree is 2.5 x 10^3 J/(kg C^degree) and that no phase change occurs within the tree itself.

Explanation / Answer

mass of water sprayed m = 7.19 kg at temperatur 0 0C

a) heat released by the water when it freezes is changing state from liq to solid so we can use

   q = m*L
      = 7.19 kg *334 KJ/kg

      = 2401.46 KJ
b)
  
   using 2401.46 KJ ,

   Q = m*C*(delta T)
     delta T = Q/m*C
         = 2401.46*10^3/(182*2.5*10^3)
         = 5.278 degrees C

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