Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W
Answer: stigmatism and discrimination
Explanation:
Knowledge of genetic risks can lead to potential social and psychological consequences for the individual. Socially, knowledge from genetic tests may lead to stigmatization and discrimination within the community.
Answer:
T1 = 625.54 K
Explanation:
We are given;
T_α = Tsur = 25°C = 298K
h = 20 W/m².K,
L = 0.15 m
K = 1.2 W/m.K
ε = 0.8
Ts = T2 = 100°C = 373K
T1 = ?
Assumption:
-Steady- state condition
-One- dimensional conduction
-No uniform heat generation
-Constant properties
From Energy balance equation;
E°in - E°out = 0
Thus,
q"cond – q"conv – q"rad = 0
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)
Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)
Thus;
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0
This gives;
(8T1 - 2984) - (1500) - 520.31 = 0
8T1 = 2984 + 1500 + 520.31
8T1 = 5004.31
T1 = 5004.31/8
T1 = 625.54 K
Answer:
429.5 ft
Explanation:
We are told that;
1 ft = 0.305 meters
We want to convert 131 meters to ft, by proportion we have;
Distance = (131 × 1)/0.305
Distance = 429.5082 ft
We want to approximate to the nearest tenth of a foot.
This gives us;
Distance = 429.5 ft
