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 8.00 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 114-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

When water freezes, it releases heat which is used to protect fruit trees from frost damage. Spraying 8.00 kg of 0°C water onto a fruit tree releases 2672 kJ of heat. This heat would increase the temperature of a 114-kg tree by approximately 0.93 °C.

Step-by-step explanation:

To calculate the amount of heat released when water freezes, we need to determine the enthalpy of fusion of water, which is the amount of heat required to convert a unit mass of water from a solid to a liquid state at its melting point.

The enthalpy of fusion of water is approximately 334 kJ/kg.

To calculate the heat released when the water freezes, we can multiply the mass of water by the enthalpy of fusion:

Heat released = mass × enthalpy of fusion = 8.00 kg × 334 kJ/kg = 2672 kJ

For part (b), to calculate the change in temperature of the tree, we can use the equation:

Change in temperature = Heat transferred / (mass × specific heat capacity)

Change in temperature = 2672 kJ / (114 kg × 2.5 x 10^3 J/(kg °C)) = 0.93 °C

Therefore, the temperature of the 114-kg tree would increase by approximately 0.93 °C.

Answer 2

The temperature of the 114-kg tree would rise by approximately 9.38°C if it absorbed the heat released when 8.00 kg of water freezes.

To solve this problem, use the equation for the heat released during a phase change:

`\[Q = m \cdot L_f\]`

where:

• 
  `\(Q\)` is the heat released during the phase change (in joules, J)
• 
  `\(m\)` is the mass of the substance undergoing the phase change (in kilograms, kg)
• 
  `\(L_f\)` is the latent heat of fusion for water, which is
  `\(3.34 * 10^5\) J/kg`

(a) Calculating the heat released when 8.00 kg of water freezes:

Given mass of water
`(\(m\)) = 8.00 kg`

Latent heat of fusion
`(\(L_f\)) = \(3.34 * 10^5\) J/kg`

`\[Q = m \cdot L_f\]`

`\[Q = 8.00 \, \text{kg} * 3.34 * 10^5 \, \text{J/kg}\]`

`\[Q = 2.67 * 10^6 \, \text{J}\]`

Therefore,
`2.67 * 10^6` joules of heat are released when 8.00 kg of water freezes.

(b) Calculating the temperature rise of the tree when it absorbs this heat:

Given mass of the tree
`(\(m_{\text{tree}}\)) = 114 kg`

Specific heat capacity of the tree
`(\(C\)) = \(2.5 * 10^3\) J/(kg degree C)`

Heat released
`(\(Q\)) = 2.67 * 10^6 J`

The formula to calculate temperature change due to the absorption of heat is:

`\[Q = m_{\text{tree}} * C * \Delta T\]`

Rearranging the formula to solve for the temperature change
`(\(\Delta T\)):`

`\[\Delta T = \frac{Q}{m_{\text{tree}} * C}\]`

`\[\Delta T = \frac{2.67 * 10^6 \, \text{J}}{114 \, \text{kg} * 2.5 * 10^3 \, \text{J/(kg degree C)}}\]`

`\[\Delta T` ≈ 9.38°C

Therefore, The answer is 9.38°C