Heat energy and liquid nitrogen boil-off rate

  • #1
Asad Raza
82
3

Homework Statement



Kindly refer to part (c). The woking should be power/0.35 (what I think according to the graph). But the answer is power (that is the answer of part b) divided by 0.02

Homework Equations


I have used the ratio method simply

The Attempt at a Solution


According to part (b), the amount of heat energy needed needed per second is 4.68 Joules (answer to part b). Now the gradient of the graph indicates that the mass loss per second is 0.35g/s. So we can imply the 4.68 Joules are needed to boil off 0.35g of liquid Nitrogen in one second. Then applying ratio, we can calculate for 1g. The correct answer to this question applies the same method but uses 0.02 instead of 0.35. Why is that so? Where did it come from?
 

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Answers and Replies

  • #2
gneill
Mentor
20,947
2,892
According to part (b), the amount of heat energy needed needed per second is 4.68 Joules (answer to part b). Now the gradient of the graph indicates that the mass loss per second is 0.35g/s. So we can imply the 4.68 Joules are needed to boil off 0.35g of liquid Nitrogen in one second. Then applying ratio, we can calculate for 1g. The correct answer to this question applies the same method but uses 0.02 instead of 0.35. Why is that so? Where did it come from?
The heater just happens to produce 4.68 W, but it's not entirely responsible for the slope of the graph line corresponding to when the heater is on; You're forgetting that even without the heat switched on, the liquid nitrogen still boils off at some rate (that's why there are two graph lines).

You need to disentangle the information given and determine what the heater's own contribution is to the mass loss rate.

One other thing. Your thread title is to generic. "On Energy" does not adequately describe the question being posed so that helpers can identify the area of physics involved. I've changed the thread title to: Heat energy and liquid nitrogen boil-off rate
 

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