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The graph is in the attachment!!! ( dont have to download to view the file)
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?
Q=m_{1}c_{1}Δt (c is specific heat capacity and Δt is change in temperature)
Q=mL_{f}
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=mL_{f} is for the amount of energy it takes to melt the ice.
Q=m_{1}c_{1}Δt is for the amount of energy it takes to heat the ice.
Q=m_{1}c_{1}Δt
=(0.25)(2.1 X 10^{3})(t_{2}t_{1})
=(0.25)(2.1 X 10^{3})(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 10^{3})(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=mL_{f} is for the amount of energy it takes to melt the ice.
=(0.25)(3.3 X 10^{5})
Q = 82500 J (it takes this much energy to melt the ice)
P=Q/Δt
Q/P = Δt
82500/105 = Δt
785.7 s =Δt
The question asks how much additional time it will take AFTER 150s, so we subtract 150 from 785.7s
785.7150
=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 dont have to download to view 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=m_{1}c_{1}Δt (c is specific heat capacity and Δt is change in temperature)
Q=mL_{f}
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=mL_{f} is for the amount of energy it takes to melt the ice.
Q=m_{1}c_{1}Δt is for the amount of energy it takes to heat the ice.
Q=m_{1}c_{1}Δt
=(0.25)(2.1 X 10^{3})(t_{2}t_{1})
=(0.25)(2.1 X 10^{3})(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 10^{3})(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=mL_{f} is for the amount of energy it takes to melt the ice.
=(0.25)(3.3 X 10^{5})
Q = 82500 J (it takes this much energy to melt the ice)
P=Q/Δt
Q/P = Δt
82500/105 = Δt
785.7 s =Δt
The question asks how much additional time it will take AFTER 150s, so we subtract 150 from 785.7s
785.7150
=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 dont have to download to view file :) )
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