So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
brainly.com/question/14548124
Answer:
Explanation:bjbhfcvvjkkknbnnnm
Explanation:
Given that,
Weight of water = 25 kg
Temperature = 23°C
Weight of mass = 32 kg
Distance = 5 m
(a). We need to calculate the amount of work done on the water
Using formula of work done



The amount of work done on the water is 1568 J.
(b). We need to calculate the internal-energy change of the water
Using formula of internal energy
The change in internal energy of the water equal to the amount of the work done on the water.


The change in internal energy is 1568 J.
(c). We need to calculate the final temperature of the water
Using formula of the change internal energy





The final temperature of the water is 23.01°C.
(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.
The amount of heat is 1568 J.
Hence, This is the required solution.
Its A The paper clip is repelled away from the nail because an electromagnetic field magnetized to the nail
Answer:

Explanation:
Given that,
Mass, m = 27 grams
Volume of the substance, V = 15 cm³
We need to find the density of the substance. We know that, the density of an object is given by mass per unit volume. So,

So, the density of the substance is equal to
.