Exercises For Section 3.1 1. Heat capacity is measured in units of J/g C, or the

Question

Exercises For Section 3.1 1. Heat capacity is measured in units of J/g C, or the energy required to raise I g of a substance by one degree Substance Heat capacity Density Air Concrete 0.88 2.4 g/cm3 Water 1.01 0.0013 g/em3 4.18 1.0 g/cm It takes 2260J to evaporate 1.0g of water (a) How much energy would it take to evaporate 1.0m2 of water 1.5 em deep? (b) How much could this energy heat the water if it didn't boil? (c) How much could this energy heat a 1.5 m thick slab of concrete with the same area? Does this match the value in the text? (d) How much could this energy heat a 10.0 km column of air? Does this match the value in the text?

Explanation / Answer

(a) Depth of water = 1.5 cm = (1.5 cm)*(1 m/100 cm) = 0.015 m.

Volume of water = (area of water)*(depth of water) = (1.0 m2)*(0.015 m) = 0.015 m3

We know that 1 m = 102 cm; therefore, 1 m3 = (102 cm)3 = 106 cm3

Therefore, 0.015 m3 = (0.015 m3)*(106 cm3/1 m3) = 15000 cm3.

Mass of water = (volume of water)*(density of water) = (15000 cm3)*(1.0 g/cm3) = 15000 g.

It takes 2260 J energy to evaporate 1.0 g water; therefore, the amount of heat taken to evaporate 15000 g water = (15000 g)*(2260 J/1 g) = 33900000 J = (33900000 J)*(1 kJ/1000 J) = 33900 kJ (ans).

(b) The water is assumed to not evaporate; rather the water shows a rise in temperature. The mass of water is 15000 g.

Apply the principle of thermochemistry.

Heat Energy supplied = (mass of water)*(specific heat of water)*(change in temperature of water)

====> 33900000 J = (15000 g)*(4.18 J/g.°C)*(change in temperature)

====> 33900000 J = (62700 J/°C)*(change in temperature)

====> change in temperature = (33900000 J)/(62700 J/°C) = 540.6698°C 540.67°C (ans).

(c) The volume of the slab of concrete = (area of the concrete)*(thickness of the concrete) = (1.0 m2)*(1.5 m) = 1.5 m3 = (1.5 m3)*(106 cm3/1 m3) = 1.5*106 cm3.

Mass of the concrete = (1.5*106 cm3)*(2.4 g/cm3) = 3.6*106 g.

Let the change in temperature be t; therefore,

33900000 J = (3.6*106 g)*(0.88 J/g.°C)*t

====> 33900000 J = (3.168*106 J/°C)*t

====> t = (33900000 J)/(3.168*106 J/°C) = 10.7°C (ans).

(d) We shall assume the area to remain the same; the length is now 10 km = (10 km)*(1000 m/1 km) = 10000 m.

Volume of air = (area of air)*(length of column) = (1.0 m2)*(10000 m) = 10000 m3 = 104 m3 = (104 m3)*(106 cm3/1 m3) = 1010 cm3.

Mass of air = (volume of air)*(density of air) = (1010 cm3)*(0.0013 g/cm3) = 1.3*107 g.

Let the change in temperature be t’; therefore,

33900000 J = (1.3*107 g)*(1.01 J/g.°C)*t’

=====> 33900000 J = (1.313*107 J/°C)*t’

=====> t’ = (33900000 J)/(1.313*107 J/°C) = 2.58°C (ans).

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