Question

At noon on a clear day in​ midsummer, a cylindrical titanium plate is placed in the sun. The plate is painted flat black so that it will absorb most of the energy from the sunlight rather than reflecting it. The plate is 5 centimeters​ [cm] in​ diameter, one centimeternbsp​[cm] ​thick, has a specific gravity of 4.5​, a molecular weight of 47.9 grams per mole​ [g/mol] and has a specific heat capacity of 25 joules per mole kelvin​ [J/(mol K)]. If the temperature of the plate increased by 17 degrees Fahrenheit​ [°F], how much energy in units of calories​ [cal] did it absorb from the​ sun, assuming no heat is reradiated to the​ surroundings?

Answer 1

Answer:

103.78 Calories

Explanation:

Given:

Diameter of the plate = 5 cm

Radius of the plate = $\frac{\textup{5}}{\textup{2}}$ = 2.5 cm

Thickness, t = 1 cm

Specific gravity = 4.5

Molecular weight = 47.9 grams/mol

specific heat capacity = 25 J/mol.K

Increase in temperature = 17° F = 17/1.8 = 9.44° C = 9.44 K

Volume of plate = πr²h = π(2.5)²(1) = 19.635 cm³

density of water = 1 g/cm³

therefore,

Density of titanium = specific gravity × density of water

or

Density of titanium = 4.5 × 1 = 4.5 g/cm³

Mass of titanium = Density × Volume

= 4.5 g/cm³ × 19.635 cm³

= 88.3575 g

Number of moles of titanium present = $\frac{\textup{Mass}}{\textup{Molar mass}}$

= $\frac{\textup{88.3575 g}}{\textup{47.9 g/mol}}$

= 1.84 moles

Energy absorbed by plate

= Number of moles × Specific heat × change in temperature

= 1.84 × 25 × 9.44

= 434.24 J

now,

1 calories = 4.184 J

or

1 J = $\frac{\textup{1}}{\textup{4.184}}$ calories

therefore,

434.24 J = $\frac{\textup{434.24}}{\textup{4.184}}$ calories

= 103.78 Calories