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 6.17 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 158-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.

Answer 1

Recall that when you are dealing with a phase change:
Q = (mL)_ice or heat transfer = (mass_ice) * (latent heat_ice)

We know that the latent heat of fusion (the change of liquid to a solid state) is 33.5E4 (J/kg)

Q = 6.17 * 33.5E4
Which tells us that 6.17kg of water freezing into its fusion state gives us:
= 2066950 J of energy.

Given a tree of 158kg and 2066950 J of energy...
With no phase change occuring:
Q = m*c*delta T
or
(heat transfer)_ice = (mass_tree * specific heat_tree * temp diff)

Q_ice = (m * c * T_final - T_inital)_tree -> where T_inital is given as 0
Q_ice = (m * c * T_final)_tree -> solve for T_final
T_f = Q_ice / (mc)_tree

Heat capacity of tree is given at 2.5E3 J/kg
T_f = (2066950 J) / (158kg)(2.5E3 J/kg)
T_f = 5.2 C degree change

*Wand drop*
Dumble Door out.