#1**+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**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**+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**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**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

#7**+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**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**+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

#10**+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