Calculating Energy Required to Heat and Change State of Ice in an Oven

[srose9625]

A student takes 0.45kg of ice. The ice is initally at -23'C. She heats the sample in an oven until the temp is 134'C.
a). What is the Q needed to heat the ice to its melting point?
b). What is the Q needed to change the solid ice into liquid water?
c). What is the Q needed to heat the liquid water to its boiling point?
d). What is the Q needed to change the liquid water to steam?
e). What is the Q needed to heat the steam to 134'C?
f). Draw a graph with the temp of the sample in Kelvin on y axis, and NRG absorbed in Joules by the sample on the x axis.


Equations of use:
Q=mc(change in T)
Q=mL

a) (.45kg) (4.2x10^3 J/kg) (23'C)=4.3x10^6 J
b) (.45kg) (3.3x10^5 J/kg) = 1.5x10^5 J
c). (.45kg) (3.3x10^5 J/kg) (100'C) = 1.2x10^7 J
I am guessing that water boils at 100'C. Would that be the right answer?
d) (.45kg) (2.3x10^6 J/kg) = 1.0x10^6 J
e). (.45kg) (3.3x10^5 J/kg) (134'C) = 1.2x10^7 J
OR
(.45kg) (2.3x10^6 J/kg) (134'C) = 1.4x10^8 J
f) I have ice drawn in a straight/horizontal line, then water drawn in a linear line followed by a straight/horizontal line, then steam in a linear line, then boiling in a straight/horizontal line.

 

[rock.freak667]

I think for part a) you need to find the heat req'd to raise the temp from -23'C to 0'C

 

[srose9625]

I think that what I did, not sure though.

 

[hage567]

Be careful with what values you use for your heat capacity "c". It is different for ice, liquid water, and steam. I think you have some mixed up.
In part (c) you have used the latent heat of fusion "L" as your heat capacity. That is wrong since you are now increasing the temperature of liquid water, not undergoing a phase change. Look up the proper values in your book.

 

[mgb_phys]

It also helps to draw a little diagram with arrows between each step and work out if the step involves taking in/giving out heat, rather than just relying on getting all the signs correct in your head.

 

[srose9625]

so for c) it would be (.45kg) (2.0x10^3 J/kg) (100'C) = 9.0x10^4 J
this is using the specific heat of steam.
Would that be right?

 

[hage567]

so for c) it would be (.45kg) (2.0x10^3 J/kg) (100'C) = 9.0x10^4 J
this is using the specific heat of steam.
Would that be right?

No, you need the specific heat for water. The water is being brought up to 100 degrees, but is not yet steam so you can't use the specific heat for steam.

 

[srose9625]

c) (.45kg) (4186 J/kg) (100'C) = 1884x10^2 J


When will I use latent heat vs. specific heat?

How did I do for e , f ?

 

[hage567]

For (e) you've got the wrong "c" value again. You need the one for steam and again you have used latent heat of fusion.

Actually, you are using the wrong one in part (a) as well. You need the one for ice.

For (f), you've sort of got the right idea, but not quite there. Think about what is happening to the ice. It is initially at -23 and is being heated to 0. So in that time, you are adding energy and it is causing the temperature to increase. So what will that look like on your graph?
The next step is the phase change from ice at 0 degrees to water at 0 degrees. What do you think this should look like? If you understand these two you can draw the rest of the graph.

 

[srose9625]

a) (.45kg) (2090 J/kg) (23'C) = 2163x10 J
e) (.45kg) (2010 J/kg) (134'C) = 1212x10^2 J
f) I have a linear line for -23'C then I have ice drawn in a straight/horizontal line, then water drawn in a linear line followed by a straight/horizontal line, then steam in a linear line, then boiling in a straight/horizontal line.
Is that right?

 

[hage567]

a) (.45kg) (2090 J/kg) (23'C) = 2163x10 J
e) (.45kg) (2010 J/kg) (134'C) = 1212x10^2 J
f) I have a linear line for -23'C then I have ice drawn in a straight/horizontal line, then water drawn in a linear line followed by a straight/horizontal line, then steam in a linear line, then boiling in a straight/horizontal line.
Is that right?

(a) looks good.

For (e), it should look like (0.45 kg)(2010 J/kg)(134-100 degrees C)
Remember, it's the change in temperature. The temperature is going from 100 to 134, so a temperature change of 34 degrees. The way you had it was how much energy would be needed to raise the temperature of steam by 134 degrees, not to 134 degrees. I should have caught that earlier.

For your graph, you should have 5 sections. You are OK up until the end. After you heat the steam to 134 that's it. There is no boiling after that. I'm not sure what you're thinking there.

 

[srose9625]

Thank you for all your help and FAST, easy to understand responses! I will chat with you again sometime:)

 

[hage567]

You're welcome.