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 6.59 kg of water at 0°C onto a fruit tree. (a) How much heat is released by the water when it freezes? (b) How much would the temperature of a 123-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 103 J/(kg C°) and that no phase change occurs within the tree itself.

Explanation / Answer

Since there is a phase change, this is a heat of fusion problem. Heat Released = mass of water x heat of fusion of water The heat of fusion of water is 334 kJ/kg , so: Heat released = 2201.06 kJ If the tree absorbed all of that heat, then we use the following equation change in energy = mass x specific heat capacity x change in temperature So (be careful of units for energy): 2201060 J = 123 kg x 2.5x10^3 J/kg-C x change in temperature The change in temperature is 7.157 C.

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