A certain kind of board transmits 3,6 x 10⁵J per day through an area of 1 m² when the temperature gradient is 1 K/cm. How many joules are transmitted per day through a board having dimensions 0,75 x 1,80 m and 4 cm thick, if one face is at 0°C and the other at 18°C?
Answers
Explanation:
Since we have a square field of side 10 meter, and one region is a smaller square of side 1 meter, there would be 100 such regions in the entire field.
Since the analysis is available for 10 regions, it would be a good approximation to multiply the total number in these ten regions by 10 to get the total number of earthworms in the field.
We thus add up the obtained figures.
⇒107+147+146+135+149+141+150+154+176+166=1471
⇒1471×10=14710, can be approximated to an order of 15000.
Answer:
The transfer of heat energy is defined as heat flux, Q. By definition, this is the flow of heat energy through a defined area over a defined time. So, the units for Q are Joules (energy) divided by area (square meters) and time (seconds). Joules/(m^2∙sec). Since power is defined as energy divided by time and 1 Watt is equal to 1 Joule/second, Q can also be expressed as Watts/m^2 .
Thermal Conductivity: The rate at which heat penetrates through a given material.
Fourier’s Law defines the ability of a specific material to transfer heat. It defines the heat flux, Q, through a material in terms of the cross-sectional area through which the heat energy transfer occurs, A, and the temperature gradient over which the transfer occurs, ∆T/∆x:
Q=-κA ∆T/∆x (1)
k is the proportionality constant for this relationship, known as the thermal conductivity. It is a characteristic property of a material. The units of the individual components of Equation (1) can be written out (with Q defined in terms of power):
Watts/m^2 =κ∙m^2∙K/m (2)
When the units of length are cancelled out, Equation (2) becomes:
Watts/m=κ∙K (3)
Solving for the thermal conductivity, k:
κ=Watts/(m∙K) (4)
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