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.
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.
To calculate the amount of heat released when the water freezes, we need to use the heat of fusion formula:
Heat released = mass of water × heat of fusion of water
(a) The mass of water is given as 6.17 kg, and the heat of fusion of water is 3.34×105 J/kg. Therefore,
Heat released = 6.17 kg × 3.34×105 J/kg
= 2.0618×106 J
So, when the water freezes, approximately 2.0618×106 J of heat is released.
(b) To calculate the temperature rise of the tree, we can use the specific heat capacity formula:
Heat = mass of tree × specific heat capacity of tree × change in temperature
Here, the mass of the tree is given as 158 kg, the specific heat capacity of the tree is 2.5×103 J/(kg C°), and the heat released is 2.0618×106 J.
Therefore,
2.0618×106 J = 158 kg × 2.5×103 J/(kg C°) × change in temperature
Solving for the change in temperature:
change in temperature = 2.0618×106 J / (158 kg × 2.5×103 J/(kg C°))
= 5.176°C
So, the temperature of the 158-kg tree would rise by approximately 5.176°C if it absorbed the heat released by the freezing of the water.
To find the answers, we need to use the specific heat capacity and the heat of fusion of water.
(a) The heat released when water freezes can be calculated using the heat of fusion equation:
Q = m * ΔHf
Where:
Q is the heat released
m is the mass of water
ΔHf is the heat of fusion of water
The heat of fusion for water is 334,000 J/kg. So, we can calculate the heat released as:
Q = 6.17 kg * 334,000 J/kg
Q ≈ 2,062,780 J
Therefore, the heat released when the water freezes is approximately 2,062,780 J.
(b) To find the temperature rise of the tree, we can use the specific heat equation:
Q = m * c * ΔT
Where:
Q is the heat absorbed by the tree
m is the mass of the tree
c is the specific heat capacity of the tree
ΔT is the temperature change
We are given the mass of the tree (158 kg) and the specific heat capacity (2.5 x 10^3 J/(kg C°)). We already found the heat released (2,062,780 J). Rearranging the equation, we can solve for ΔT:
ΔT = Q / (m * c)
ΔT = 2,062,780 J / (158 kg * 2.5 x 10^3 J/(kg C°))
ΔT ≈ 5.20 °C
Therefore, the temperature of the tree would rise approximately 5.20 °C if it absorbed the heat released during the freezing of the water.