Problem 52 - Physics

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 \(7.2 \mathrm{~kg}\) of water at \(0^{\circ} \mathrm{C}\) onto a fruit tree. (a) How much heat is released by the water when it freezes? (b) How much would the temperature of a \(180-\mathrm{kg}\) tree rise if it absorbed the heat released in part (a)? Assume that the specific heat capacity of the tree is \(2.5 \times 10^{3} \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and that no phase change occurs within the tree itself.

Step-by-Step Solution

Answer

(a) 2,404,800 J; (b) 5.34°C

Step 1: Identify the Heat Released when Water Freezes

The latent heat of fusion (L_f) for water is the amount of heat released when 1 kg of water freezes. It is given by L_f = 334,000 \text{ J/kg}. For 7.2\text{ kg} of water, the heat (Q) released when water turns to ice is calculated using:\[Q = m \times L_f\]Substitutem = 7.2 \text{ kg} and L_f = 334,000 \text{ J/kg} into the formula:\[Q = 7.2 \text{ kg} \times 334,000 \text{ J/kg} = 2,404,800 \text{ J}\]

Step 2: Determine the Temperature Rise of the Tree

The heat absorbed by the tree will cause a temperature rise. This can be calculated using the formula:\[Q = mc\Delta T\]where Q is the heat absorbed by the tree, m is the mass of the tree (180 \text{ kg}), c is the specific heat capacity of the tree (2.5 \times 10^3 \text{ J/(kg} \cdot \text{C}^{\circ})), and \Delta T is the temperature change. Re-arranging for \Delta T gives:\[\Delta T = \frac{Q}{mc}\]Substitute Q = 2,404,800 \text{ J}, m = 180 \text{ kg}, and c = 2.5 \times 10^3 \text{ J/(kg} \cdot \text{C}^{\circ}):\[\Delta T = \frac{2,404,800 \text{ J}}{180 \text{ kg} \times 2.5 \times 10^3 \text{ J/(kg} \cdot \text{C}^{\circ})} = 5.34 \text{ C}^{\circ}\]

Key Concepts

Latent Heat of Fusion

When water freezes, it undergoes a phase change from liquid to solid. During this process, heat is released despite the temperature remaining constant. This is known as the latent heat of fusion.

• Definition: Latent heat of fusion (\(L_f\) is the energy required to change 1 kg of a solid into a liquid without changing its temperature. For water, \(L_f\) is 334,000 J/kg.
• Significance in the Problem: In the given exercise, 7.2 kg of water is sprayed on crops. As it freezes, it releases heat, providing warmth to the plants.
• Calculation: The total heat released (Q) can be calculated using the formula: \[Q = m \times L_f\]where \(m\) is the mass of water (7.2 kg). Thus:\[Q = 7.2 \text{ kg} \times 334,000 \text{ J/kg} = 2,404,800 \text{ J}\]

This heat helps in preventing frost damage by maintaining a slightly warmer temperature around the plants.

Specific Heat Capacity

Specific heat capacity is a measure of the amount of heat energy required to raise the temperature of a given mass of a substance by one degree Celsius.

• Definition: It is symbolized as \(c\) and usually expressed in units of \(J/(kg \cdot C^{\circ})\).
• Role in the Problem: In this exercise, the specific heat capacity (\(c\)) of the tree is given as \(2.5 \times 10^3 \text{ J/(kg} \cdot \text{C}^{\circ})\). This value tells us how much heat the tree can absorb per kilogram per degree Celsius rise in temperature.
• Application: When heat is absorbed, the temperature change (\(\Delta T\)) of the tree can be found using:\[Q = mc\Delta T\] where \(Q\) is the absorbed heat, and \(m\) is the mass of the tree (180 kg). Rearranging the equation gives: \[\Delta T = \frac{Q}{mc}\]
• Conclusion: By substituting the values, we see that the temperature rise of the tree is \(5.34 \text{ C}^{\circ}\).

This concept reflects how heat capacity impacts temperature changes in materials.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or substance to another. In this scenario, it is the process by which the released heat from freezing water is transferred to the tree.

• Components: Heat transfer can occur via conduction, convection, or radiation. Here, conduction is primarily at play, as heat moves through direct contact from water-turned-ice to tree tissue.
• Thermodynamic Implication: Heat transfer follows the second law of thermodynamics, indicating that energy naturally flows from a hotter body to a cooler one until equilibrium is reached.
• Problem Application: The released energy of \(2,404,800 \text{ J}\) from the freezing water is transferred to the tree, increasing its temperature and offering frost protection. The energy applied to the tree increases its environmental temperature, providing protective warmth against overnight freezing conditions.

Understanding heat transfer is crucial for explaining how the heat released during water's phase change to ice supports the tree's defense against cold.