How long to melt the ice(heat, heat transfer, latent heat)

In summary, the conversation discusses solving a problem involving heating and melting a piece of ice using equations for specific heat capacity and energy. The graph provided shows the temperature change over time and the question asks how much additional time will be required to melt all of the ice after 150 seconds, assuming a constant power from the heater. The solution involves calculating the energy required for heating and melting, determining the power of the heater, and solving for the additional time needed to melt the ice. The methodology used in the solution is deemed correct, although some adjustments may need to be made for the specific temperature changes shown on the graph.
  • #1
supernova1203
210
0
The graph is in the attachment! ( don't have to download to view the file)

Homework Statement


49a) A 0.25 kg piece of ice is warmed by an electric heater and the following graph of temperature is produced. Assume that there has been no loss of energy to surroundings.

How much additional time after 150s will be required to melt all of the ice, assuming the power on the heater is constant?

Homework Equations



Q=m1c1Δt (c is specific heat capacity and Δt is change in temperature)

Q=mLf

The Attempt at a Solution

Another way to word the question 49a would be to ask ‘ How long does it take for ice to get from -10 degrees to 0 degrees

Q=mLf is for the amount of energy it takes to melt the ice.Q=m1c1Δt is for the amount of energy it takes to heat the ice.Q=m1c1Δt

=(0.25)(2.1 X 103)(t2-t1)

=(0.25)(2.1 X 103)(0-(-30) <------the ice is initially at -30 as you can see on the graph.(graph is in attachment, and can be viewed without downloading)
=(0.25)(2.1 X 103)(0+30)

Q=15750 J

This is the amount of energy required to heat the ice.
P = W/Δt (Δt is now for time)since W = ΔE = Q P= 15750 J/150s (this 150s was given, as you can see on graph)

P = 105 W ( this remains constant as the question states)Q=mLf is for the amount of energy it takes to melt the ice.

=(0.25)(3.3 X 105)

Q = 82500 J (it takes this much energy to melt the ice)
P=Q/ΔtQ/P = Δt

82500/105 = Δt785.7 s =ΔtThe question asks how much additional time it will take AFTER 150s, so we subtract 150 from 785.7s

785.7-150
=635.7s It will take an additional 635.7s to melt the ice.

or 635.7/60

= 10.5 mins or it will take 10.5 mins to melt the ice.Does this look right? If not can you show me how to get the correct solution? Or better yet, if i have the incorrect solution, get the right solution and show it to me and i will figure out on my own on how to get the right solution.

Thanks :)

(graph is in the attachment, as always you don't have to download to view file :) )
 

Attachments

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  • #2
For the time interval shown on the graph (150s), the ice temperature change is not 30C. It starts at -30C and rises to -10C over the 150s. This will change your heater power calculation and subsequent values.
 
  • #3
gneill said:
For the time interval shown on the graph (150s), the ice temperature change is not 30C. It starts at -30C and rises to -10C over the 150s. This will change your heater power calculation and subsequent values.

ahh... i see

So...everything else looks good? the procedure and the steps i have and the method i used to get the solution is fine?


Also the reason i put it to 0 degrees instead of -10 is because i always understood t2 was usually the melting point
 
  • #4
supernova1203 said:
ahh... i see

So...everything else looks good? the procedure and the steps i have and the method i used to get the solution is fine?
Your methodology looks okay; you seem to understand the required stages for solution.
Also the reason i put it to 0 degrees instead of -10 is because i always understood t2 was usually the melting point
A variable is what you say it is :smile: When analyses become more complex than "generic examples" you have to be flexible in your variable assignments :smile:
 
  • #5


Your solution looks correct to me. You correctly used the equations for specific heat capacity and latent heat to calculate the amount of energy required to heat and melt the ice. You then used the equation for power to calculate the time it would take to melt the ice, and accounted for the initial 150s given in the question. Your final answer of 10.5 minutes to melt the ice is also correct. Great job!
 

1. How does heat affect the melting of ice?

Heat is the main factor that causes ice to melt. When heat is applied to ice, the molecules within the ice start to vibrate and move faster, causing the ice to break apart and turn into liquid water.

2. What is the process of heat transfer during ice melting?

The process of heat transfer during ice melting is called conduction. This is when heat is transferred from a warmer object to a cooler object through direct contact. In the case of ice, heat is transferred from the warmer environment to the cooler ice, causing it to melt.

3. What is latent heat and how does it relate to ice melting?

Latent heat is the energy required to change the state of a substance without changing its temperature. In the case of ice melting, the latent heat of fusion is the amount of energy needed to convert the ice into liquid water without changing its temperature. This is why ice can remain at 0 degrees Celsius while it is melting.

4. How long does it take for ice to melt at different temperatures?

The time it takes for ice to melt depends on various factors such as the temperature of the environment, the size and shape of the ice, and the amount of heat applied. Generally, it takes around 2-3 minutes for a small ice cube to melt at room temperature.

5. Can other factors besides heat affect the melting of ice?

Yes, other factors such as pressure and impurities can also affect the melting of ice. Higher pressure can lower the melting point of ice, while impurities in the ice can disrupt the molecular structure and make it more difficult for the ice to melt.

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