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 6.84 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 117-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

Heat from freezing (Q) is the mass of the water (m) times the heat of fusion.

Q = 6.84 kg * 334 kJ/kg

Q = 2284.5 kJ

For the tree absorbing the heat, the specific heat capacity (C) = the heat absorbed (Q) / the temperature difference (T2-T1)

C = Q / (T2 - T1)

T2 - T1 = Q / C

T2 = Q / C + T1

But they didn't give us C, the gave use the SPECIFIC heat capacity (Cm), the heat capacity per Mass of tree (M).

Cm = C / M

C = Cm * M

Plug that into our other equation

T2 = Q / C + T1

T2 = (Q / (Cm * M)) + T1

T2 = 2284500 J / (( 2500 J/(kg C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.