+36916 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
+37 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
+36916 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!
+2440 EP, the heat of fusion for 1g of water at 0 °C is approximately 334 joules or 79.7 calories.
GA
+37 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?
+36916 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?
+37 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.
+36916 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.
+36916 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.
