Thermodynamics question with heat gain from surroundings considered

AI Thread Summary
The discussion centers on calculating the latent heat of fusion of ice and the heat gained from surroundings using two methods based on experimental data from a heater melting ice. Method 1 involves using the energy supplied by the heater and the heat gained from the surroundings, leading to a calculated latent heat of 333 J/g and heat gain of 16.7 J/s. Method 2, which calculates latent heat based solely on heater output, yields inconsistent results due to neglecting the heat gained from the surroundings. Participants emphasize that Method 1 is more accurate, while Method 2's approach leads to contradictory answers. Clarifications are sought regarding the assumptions made in Method 1 about heat loss from the heater.
FaroukYasser
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Homework Statement


A heater with a variable power is surrounded by ice in a funnel. The heater is turned on on 70W and the ice starts melting. once the rate of dropping of droplets is constant. The mass of water falling is then calculated for 5 minutes and it is found that 0.26 g/s falls every second. The experiment is repeated for 110W for 5 minutes also and the mass is found to be 0.38 g/s, Find:

1) The latent heat of fusion of ice in J/g
2) The heat gained from the surrounding per second

Homework Equations


Energy supplied by heater + heat gained from surrounding = mass * latent heat of fusion or in symbols:
E + h = mL

The Attempt at a Solution



Okay so I am having two ideas to solve this and I am not sure which one I should choose:
Method 1:
For the first time with 70 W.
E + h = mL
(70x1) + h = 0.26L (Equation 1)

For the second time with 110 W
(110x1) + h = 0.38L (Equation 2)

Equation 2 - Equation 1
40 = 0.12L >>>> L = 333 J/g to 3s.f

subbing in L into any equation. h = 16.7 J/s (W)

Method 2
L = E/m
using the first value once. L = 70/0.26 = 269.2
using second valiue of L = 110/0.38 = 289.5
Average is 279 J/gWhich method should I use?. does method one have any flaws and if so what are they. does Method 2 have any flaws and if so what is it?

Thanks in advance :)
 
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Your 1st method looks right. The wording is poor. There is heat gained from the surroundings plus heat gained from the heater.
Your 2nd method ignores heat h from the surroundings so you get two contradictory answers.
 
rude man said:
Your 1st method looks right. The wording is poor. There is heat gained from the surroundings plus heat gained from the heater.
Your 2nd method ignores heat h from the surroundings so you get two contradictory answers.
Hi
Thanks for your reply :). I was wandering what you meant by "wording is poor" and if there is anything i can improve. Also in method 1, are we neglecting the heat energy that is lost from the heater to the surrounding or does the constant rate of dripping indicate that all the heat from the heater is supplied to the ice?
Thanks in advance :)
 
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