An open container holds ice of mass 0.500 kgkg at a temperature of -17.4 âCâC . The mass of the container can be ignored. Heat i
2 answers:
Answer:
It takes 25.73 minutes before the ice starts to melt, and it takes 235.21 minutes (about 3.9 hours) before the temperature of ice begins to rise above 0°C.
Explanation:
First, we need to know how much heat is required to rise the temperature of ice to 0°C. To do that, we use the following equation (using the equivalence -17.4°C≈255.6°K and 0°C≈273°K):
So are needed 18,720 Joules of heat. Since there is supplied heat at a constant rate of 710 J/min, we can get the time that passes before the ice startes to melt from the equation:
Then, it passes 25.73 minutes before the ice starts to melt.
Now, we do the same to get the time it takes before the temperature begins to rise above 0°C :
Then, are needed 167,000 Joules of heat to completely melt ice. Since a substance in a phase change does not change its temperature, when ice receives this amount of heat then it starts to rise its temperature above 0°C. So, the time is obtained by:
Finally, it passes 235.21 minutes (about 3.9 hours) before the temperature of ice begins to rise above 0°C.
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Answer:
The answers to the question are
(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W
(ii) The temperature of the outer surface of the insulation is 49.89 °C
Explanation:
To solve the question, we note that the heat transferred is given by
Where
= Temperature at the inside of the pipe = 300 °C
= Temperature at the outside of the pipe = 20 °C
r₁ =internal radius of pipe = 4.0 cm
r₂ = Outer radius of pipe = 4.5 cm
r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm
= 15 W/m·K
= 0.038 W/m·K
= 75 W/m²·K
= 10 W/m²·K
Plugging in the values in the above equation where for a unit length L = 1 m, we have
Q = 131.32 W
From which we have, for the film of air at the pipe outer boundary layer
Where
for the air film on the pipe outer surface is given by
where A =area of the outside of the pipe
= = 0.227 K/W
Therefore
131.32 W = which gives
= 49.89 °C
Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)
Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)
T₁ = Surface temperature of the pipe = 49.89 °C and
T₂ = Temperature of the surrounding = 20.00 °C
Plugging in the values gives, q' = 0.307 W per m²
Total heat lost per unit length = 131.32 + 0.307 =131.62 W
Answer:
Explanation:
Given, for girl : Weight or force;
Area of both heels;
For elephant, Weight = Force = 2000 kg•f
Area of 4 feet;
Now;
Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.
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