About Sugar
Sugar is colourless, water-soluble compound present in the sap of seed plants and the milk of mammals and making up the simplest group of carbohydrates. The most common sugar is sucrose, a crystalline tabletop and industrial sweetener used in foods and beverages.
Summary
Name | Sugar |
Phase at STP | solid |
Density | 1600 kg/m3 |
Ultimate Tensile Strength | N/A |
Yield Strength | N/A |
Young’s Modulus of Elasticity | N/A |
Brinell Hardness | N/A |
Melting Point | 186 °C |
Thermal Conductivity | 0.15 W/mK |
Heat Capacity | 1244 J/g K |
Price | 1 $/kg |
Density of Sugar
In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m3). The Standard English unit is pounds mass per cubic foot (lbm/ft3).
Density of Sugar is 1600 kg/m3.
Example: Density
Calculate the height of a cube made of Sugar, which weighs one metric ton.
Solution:
Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V
As the volume of a cube is the third power of its sides (V = a3), the height of this cube can be calculated:
The height of this cube is then a = 0.855 m.
Density of Materials
Thermal Properties of Sugar
Sugar – Melting Point
Melting point of Sugar is 186 °C.
Note that, these points are associated with the standard atmospheric pressure. In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.
Sugar – Thermal Conductivity
Thermal conductivity of Sugar is 0.15 W/(m·K).
The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.
The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:
Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.
Sugar – Specific Heat
Specific heat of Sugar is 1244 J/g K.
Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties cv and cp are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:
where the subscripts v and p denote the variables held fixed during differentiation. The properties cv and cp are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol K.
Example: Heat transfer calculation
Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.
Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m2). The wall is 15 cm thick (L1) and it is made of Sugar with the thermal conductivity of k1 = 0.15 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h1 = 10 W/m2K and h2 = 30 W/m2K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).
Calculate the heat flux (heat loss) through this wall.
Solution:
As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:
The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.
Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:
The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/0.15 + 1/30) = 0.88 W/m2K
The heat flux can be then calculated simply as: q = 0.88 [W/m2K] x 30 [K] = 26.47 W/m2
The total heat loss through this wall will be: qloss = q . A = 26.47 [W/m2] x 30 [m2] = 794.12 W