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1. How much heat in joules is needed to raise the temperature of 7.0 L of water from 0°C to 78.0°C?

 

2. How much heat is necessary to change 300 g of ice at -15°C to water at 20°C?

 

Thank you!!

cmoore  Mar 2, 2018
edited by cmoore  Mar 2, 2018
edited by cmoore  Mar 2, 2018
 #1
avatar+13044 
+1

First let's find out how many CALORIES are required, and then convert to joules , knowing

1 cal = 4.184 j

 

One calorie will raise ONE GRAM of water ONE DEGREE Celsius

#calories = 7000 gm  x 78 C = 546000  CALORIES

 

546000 calories x 4.184 j/calorie = 2284.5 kj    (kj = KILOjoule)

 

 

2.  First you have to warm the ice up to 0 degrees

     300 g x 1cal/gram-C x  15 C = 4500 cal

 

     Then you have to add heat to it AT A CONSTANT TEMP to melt the ice @ 80 cal/gm

300gm x 80 cal/gm = 24000 cal

 

     Then you have to warm the water (melted ice at 0 degrees C) to 20 C

300 gm x 1 cal/gram-C  x 20 C  = 6000 calories

 

Add them all up  4500+24000+6000 = 34500 cal     =    144.3 kj

ElectricPavlov  Mar 2, 2018
 #2
avatar+37 
0

Unfourtanetly that wasn't correct.

I have been given this information, but I still can't seem to get the correct answer.

 

1. How much heat in joules is needed to raise the temperature of 7.5 L of water from 0°C to 87.0°C? (Hint: Recall the original definition of the liter.)
 2.73e+06 J

 

 

2.  How much heat is necessary to change 325 g of ice at -20°C to water at 20°C?

change g to kilograms; and to figure the heat here because it is original ice and will need to be melted you will do

The specific heat of ice calculation to get it from -20 to 0 °C + the heat of fusion to melt the ice +  the specific heat calculation to raise it to 20°C :

The specific heat of ice is .5kcal/kg*°C

 

H = m c ice ΔT + mLf  + m c ΔT =

 .325 kg (.5kcal   ) (20°C) + .325 kg (80 kcal)  + .325 kg ( 1 kcal)  (20°C)  = 35.75 kcal

                kg *°C                                        Kg                      kg * °C

cmoore  Mar 2, 2018
 #3
avatar+13044 
+1

I apologize.....I used the wrong specific heat for ice   I use 1     it is .5

 

so to heat the ice to zero degrees will require

300 x .5 x 15 = 2250 cal    (not 4500)

 

 

NOW add them up   2250 + 24000 + 6000 = 32250 cal = 134934 j = 134.9 kj

 

 

Sorry!

ElectricPavlov  Mar 2, 2018
 #4
avatar+1184 
0

EP, the heat of fusion for 1g of water at 0 °C is approximately 334 joules or 79.7 calories. 

 

 

 

GA

GingerAle  Mar 2, 2018
 #5
avatar+37 
0

Thank you! I tried to redo it, but I just could not figure it out.

The first answer was wrong as well, the answer needs to be in this format. 2.73e+06 J
Any ideas?

cmoore  Mar 2, 2018
 #6
avatar+13044 
+1

Thanx GA...I have always used 80  

ElectricPavlov  Mar 2, 2018
 #7
avatar+13044 
+1

Well then the first answer becomes

2284.5 kj = 2284500j = 2.28 e+06

 

I am thinking there may be a slight conversion factor error in there....or an error in your answer.....

    I see no other way to get other than the answer I calculated.

 

What numbers are you using for specific heat of water?

ElectricPavlov  Mar 2, 2018
 #8
avatar+37 
0

This problem was just an example I had. 

 

How much heat in joules is needed to raise the temperature of 7.5 L of water from 0°C to 87.0°C? (Hint: Recall the original definition of the liter.)
 2.73e+06 J

 

The one I need to answer is this. 

 

How much heat in joules is needed to raise the temperature of 7.0 L of water from 0°C to 78.0°C? (Hint: Recall the original definition of the liter.)

 

I had already attempted to enter the answer in as you just said and it didn't work. So I'm really not sure what is going wrong.

cmoore  Mar 2, 2018
 #9
avatar+13044 
+1

Yes, if I use your EXAMPLE

 

How much heat in joules is needed to raise the temperature of 7.5 L of water from 0°C to 87.0°C? (Hint: Recall the original definition of the liter.)
 2.73e+06 J

 

 

7.5 L x 1000 gm/l   1 cal/gm-c x 87c = 652500 cal = 2.73 e +6    checks out fine

    I think the answer they are giving you is incorrect....

        either the temp change is wrong or the amount of water is incorrect.  

ElectricPavlov  Mar 2, 2018
edited by ElectricPavlov  Mar 2, 2018
 #11
avatar+37 
0

Yeah I've redone it a few times and I keep getting 2.28e+06 J, and it's still saying it's wrong.

 

Thank you for all of your help. I really appreciate it. 

cmoore  Mar 2, 2018
 #13
avatar+37 
0

Apparently I've been up to long. I didn't put the 10 in when I answered the question! Thanks again for your help. smiley

cmoore  Mar 2, 2018
 #10
avatar+13044 
+1

GA  pointed out that in j the heat of fusion is 333.55 j/g  ~ 79.72 cal/g   

    I used 80 cal/gm   (which is what we used when I was in H.S. Physics many years ago)  in my calc....so you may want to change that constant to get a slightly different answer to the SECOND question.....  OK ????   Instead of 24000 it would be 23916 calories for that part of the solution.

ElectricPavlov  Mar 2, 2018
edited by ElectricPavlov  Mar 2, 2018
 #12
avatar+37 
0

Yes, the SECOND question is right. After I converted it ends up being 32.24kcal. 

cmoore  Mar 2, 2018

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